vieta的惯例据/h1>
已经有一个帐户?据一种href="//www.parkandroid.com/account/login/?next=/wiki/vietas-formula/" class="ax-click" data-ax-id="clicked_signup_modal_login" data-ax-type="link">这里登录。据/一种>据/P.>据/div>
建议课程据/h4>
-
比赛数学二据/h3>
在AMC 10和12的水平下解决数学问题的指导训练。据/P.>据/div>
测验据/h4>
相关......据/h4>
- 代数据/span>>据/span>
贡献据/div>
韦达的公式据/strong>涉及的系数据一种href="//www.parkandroid.com/wiki/polynomials/" class="wiki_link" title="多项式" target="_blank">多项式据/一种>到的款项及其根源的产品,以及在群体采取根的产品。据/P.>据P.>例如,如果存在二次多项式据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>X据/span>2据/span>+据/span>2据/span>X据/span>-据/span>1据/span>5.据/span>,则其根据span class="katex">
X据/span>=据/span>-据/span>5.据/span>和据span class="katex">
X据/span>=据/span>3.据/span>, 因为据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>X据/span>2据/span>+据/span>2据/span>X据/span>-据/span>1据/span>5.据/span>=据/span>(据/span>X据/span>-据/span>3.据/span>)据/span>(据/span>X据/span>+据/span>5.据/span>)据/span>。韦达的公式可以找到根的总和据span class="katex">
(据/span>3.据/span>+据/span>(据/span>-据/span>5.据/span>)据/span>=据/span>-据/span>2据/span>)据/span>和根的产品据span class="katex">
(据/span>3.据/span>⋅据/span>(据/span>-据/span>5.据/span>)据/span>=据/span>-据/span>1据/span>5.据/span>)据/span>而不直接找到每一根。虽然这是在这个具体的例子相当琐碎,韦达的公式是在更复杂的代数多项式有许多根或当多项式的根不容易推导是非常有用的。对于一些问题,韦达的公式可以作为快捷方式寻求解决办法很快知道自己的根的款项或产品。据/P.>据/div>
内容据/h4>
韦达的公式 - 二次方程据/h2>
让我们从定义开始。据/P.>据B.lockquote class="definition">
韦达的公式二次方程式据/strong>:据/P.>据P.>给予据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>一种据/span>X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>如果方程据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>0.据/span>有根据span class="katex">
R.据/span>1据/span>和据span class="katex">
R.据/span>2据/span>, 然后据/P.>据P.>据span class="katex-display">
R.据/span>1据/span>+据/span>R.据/span>2据/span>=据/span>-据/span>一种据/span>B.据/span>那据/span>R.据/span>1据/span>R.据/span>2据/span>=据/span>一种据/span>C据/span>。据/span>□据/span>
这句话的证明在本节的最后给出。据/P.>据P.>我们立即看到的二次帮助系数如何确定根的关系。据/P.>据B.lockquote class="example">
如果据span class="katex">
α.据/span>和据span class="katex">
β据/span>是二次的根源据span class="katex">
X据/span>2据/span>-据/span>4.据/span>X据/span>+据/span>9.据/span>=据/span>0.据/span>什么是值据/P.>据ol>
-
α.据/span>+据/span>β据/span>
-
α.据/span>β据/span>
-
α.据/span>2据/span>+据/span>β据/span>2据/span>还是据/span>
从韦达的公式,我们认识到,据span class="katex">
α.据/span>+据/span>β据/span>=据/span>4.据/span>。据/P.>据/li>
从韦达的公式,我们认识到,据span class="katex">
α.据/span>β据/span>=据/span>9.据/span>。据/P.>据/li>
韦达的公式并没有告诉我们的价值据span class="katex">
α.据/span>2据/span>+据/span>β据/span>2据/span>直接地。我们需要做的,是写的据span class="katex">
α.据/span>2据/span>+据/span>β据/span>2据/span>按照据span class="katex">
α.据/span>+据/span>β据/span>和/或据span class="katex">
α.据/span>β据/span>,然后我们可以替换这些值。我们有据/P.>据/li>
α.据/span>2据/span>+据/span>β据/span>2据/span>=据/span>(据/span>α.据/span>+据/span>β据/span>)据/span>2据/span>-据/span>2据/span>α.据/span>β据/span>=据/span>4.据/span>2据/span>-据/span>2据/span>×据/span>9.据/span>=据/span>-据/span>2据/span>。据/span>□据/span>
笔记据/strong>:在这个问题上,根是复数据span class="katex">
2据/span>±据/span>5.据/span>
一世据/span>。如果我们要将这些产品作为价值观据span class="katex">
α.据/span>和据span class="katex">
β据/span>然后计算,我们必须使粗心犯错计算的机会较高。韦达的公式我们提供了一个更简单的方法。据/P.>据!-- end-example -->
你能快速猜出二次的根源吗?据span class="katex">
X据/span>2据/span>-据/span>5.据/span>X据/span>+据/span>6.据/span>还是据/span>
如果据span class="katex">
P.据/span>和据span class="katex">
问:据/span>是给定方程的根,然后韦达的公式告诉我们据/P.>据P.>据span class="katex-display">
P.据/span>+据/span>问:据/span>=据/span>5.据/span>那据/span>P.据/span>问:据/span>=据/span>6.据/span>那据/span>
这并不难看据span class="katex">
2据/span>+据/span>3.据/span>=据/span>5.据/span>和据span class="katex">
2据/span>×据/span>3.据/span>=据/span>6.据/span>。据/span>所以根必须是据span class="katex">
2据/span>和据span class="katex">
3.据/span>,他们确实是。据span class="katex">
□据/span>
给予据span class="katex">
α.据/span>和据span class="katex">
β据/span>是二次的根源据span class="katex">
一种据/span>X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>=据/span>0.据/span>, 表达据span class="katex">
一种据/span>2据/span>B.据/span>2据/span>-据/span>4.据/span>一种据/span>C据/span>按照据span class="katex">
α.据/span>和据span class="katex">
β据/span>。据/P.>据hr>
韦达的公式给了我们据span class="katex">
一种据/span>-据/span>B.据/span>=据/span>α.据/span>+据/span>β据/span>和据span class="katex">
一种据/span>C据/span>=据/span>α.据/span>β据/span>。取代这些,我们得到了据/P.>据P.>据span class="katex-display">
一种据/span>2据/span>B.据/span>2据/span>-据/span>4.据/span>一种据/span>C据/span>=据/span>一种据/span>2据/span>B.据/span>2据/span>-据/span>4.据/span>一种据/span>C据/span>=据/span>(据/span>α.据/span>+据/span>β据/span>)据/span>2据/span>-据/span>4.据/span>α.据/span>β据/span>=据/span>α.据/span>2据/span>+据/span>2据/span>α.据/span>β据/span>+据/span>β据/span>2据/span>-据/span>4.据/span>α.据/span>β据/span>=据/span>α.据/span>2据/span>-据/span>2据/span>α.据/span>β据/span>+据/span>β据/span>2据/span>=据/span>(据/span>α.据/span>-据/span>β据/span>)据/span>2据/span>。据/span>□据/span>
你可能认识据span class="katex">
B.据/span>2据/span>-据/span>4.据/span>一种据/span>C据/span>来自据一种href="//www.parkandroid.com/wiki/quadratic-formula/" class="wiki_link" title="二次公式" target="_blank">二次公式据/一种>。事实上,我们可以表明这一点据/P.>据P.>据span class="katex-display">
2据/span>一种据/span>-据/span>B.据/span>±据/span>B.据/span>2据/span>-据/span>4.据/span>一种据/span>C据/span>
=据/span>2据/span>1据/span>(据/span>一种据/span>-据/span>B.据/span>±据/span>一种据/span>2据/span>B.据/span>2据/span>-据/span>4.据/span>一种据/span>C据/span>
)据/span>=据/span>2据/span>(据/span>α.据/span>+据/span>β据/span>)据/span>±据/span>|据/span>α.据/span>-据/span>β据/span>|据/span>=据/span>α.据/span>或者据/span>β据/span>那据/span>
这证明了二次公式。据/P.>据B.lockquote class="proof">
由这件事据一种href="//www.parkandroid.com/wiki/remainder-factor-theorem/" class="wiki_link" title="剩余因子定理" target="_blank">剩余因子定理据/一种>由于多项式据span class="katex">
F据/span>(据/span>X据/span>)据/span>有根据span class="katex">
R.据/span>1据/span>和据span class="katex">
R.据/span>2据/span>,它必须有形式据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>一种据/span>(据/span>X据/span>-据/span>R.据/span>1据/span>)据/span>(据/span>X据/span>-据/span>R.据/span>2据/span>)据/span>=据/span>一种据/span>X据/span>2据/span>-据/span>一种据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>)据/span>X据/span>+据/span>一种据/span>R.据/span>1据/span>R.据/span>2据/span>对于某一常数据span class="katex">
一种据/span>。据/P.>据P.>比较系数据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>一种据/span>X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>,我们得出结论据span class="katex">
一种据/span>=据/span>一种据/span>那据/span>B.据/span>=据/span>-据/span>一种据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>)据/span>和据span class="katex">
C据/span>=据/span>一种据/span>R.据/span>1据/span>R.据/span>2据/span>。因此,我们得到据/P.>据P.>据span class="katex-display">
R.据/span>1据/span>+据/span>R.据/span>2据/span>=据/span>-据/span>一种据/span>B.据/span>那据/span>R.据/span>1据/span>R.据/span>2据/span>=据/span>一种据/span>C据/span>。据/span>□据/span>
韦达的公式 - 成型二次方程式据/h2>
让据span class="katex">
P.据/span>和据span class="katex">
问:据/span>是首一元二次方程的真正根源据/P.>据P.>据span class="katex-display">
X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>=据/span>0.据/span>那据/span>(据/span>B.据/span>那据/span>C据/span>)据/span>∈据/span>R.据/span>2据/span>。据/span>
为什么摩尼?因为我们总是可以通过其领先的系数将整个方程来获得它的首一版本。所以,我们能说的据span class="katex">
B.据/span>那据/span>C据/span>还是自从据span class="katex">
P.据/span>和据span class="katex">
问:据/span>此方程的根,我们可以因式分解公式为据/P.>据P.>据span class="katex-display">
X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>≡据/span>(据/span>X据/span>-据/span>P.据/span>)据/span>(据/span>X据/span>-据/span>问:据/span>)据/span>。据/span>
扩大右侧并重新排列,我们发现据/P.>据P.>据span class="katex-display">
X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>≡据/span>X据/span>2据/span>-据/span>(据/span>P.据/span>+据/span>问:据/span>)据/span>X据/span>+据/span>P.据/span>问:据/span>。据/span>
由于两个多项式相等,当且仅当它们的系数是相等的,由等同我们得到的系数据/P.>据P.>据span class="katex-display">
B.据/span>=据/span>-据/span>(据/span>P.据/span>+据/span>问:据/span>)据/span>那据/span>C据/span>=据/span>P.据/span>问:据/span>。据/span>
这就是所谓的韦达公式为二次多项式。它也同样能够扩展到多项式据一种target="_blank" rel="nofollow" href="#vietas-formula-higher-degrees">程度更高据/一种>。据/P.>据P.>根可以广泛地包括复数。也就是说,给定两个复数据span class="katex">
P.据/span>和据span class="katex">
问:据/span>,我们总能构建一个首一二次,其根源是据span class="katex">
P.据/span>和据span class="katex">
问:据/span>。更具体地讲,二次会据/P.>据P.>据span class="katex-display">
X据/span>2据/span>-据/span>(据/span>P.据/span>+据/span>问:据/span>)据/span>X据/span>+据/span>P.据/span>问:据/span>=据/span>0.据/span>。据/span>
这两个系数的时候会是真的吗?为了找到答案,集据span class="katex">
P.据/span>=据/span>P.据/span>1据/span>+据/span>P.据/span>2据/span>一世据/span>和据span class="katex">
问:据/span>=据/span>问:据/span>1据/span>+据/span>问:据/span>2据/span>一世据/span>。然后系数是据/P.>据P.>据span class="katex-display">
B.据/span>=据/span>-据/span>(据/span>P.据/span>1据/span>+据/span>问:据/span>1据/span>)据/span>-据/span>(据/span>P.据/span>2据/span>+据/span>问:据/span>2据/span>)据/span>一世据/span>那据/span>C据/span>=据/span>(据/span>P.据/span>1据/span>问:据/span>1据/span>-据/span>P.据/span>2据/span>问:据/span>2据/span>)据/span>+据/span>(据/span>P.据/span>1据/span>问:据/span>2据/span>+据/span>P.据/span>2据/span>问:据/span>1据/span>)据/span>一世据/span>。据/span>
为了据span class="katex">
B.据/span>要真实,我们需要据span class="katex">
P.据/span>2据/span>+据/span>问:据/span>2据/span>=据/span>0.据/span>⇒据/span>P.据/span>2据/span>=据/span>-据/span>问:据/span>2据/span>。为了据span class="katex">
C据/span>要真实,我们需要据/P.>据P.>据span class="katex-display">
P.据/span>1据/span>问:据/span>2据/span>+据/span>P.据/span>2据/span>问:据/span>1据/span>=据/span>0.据/span>⇒据/span>问:据/span>2据/span>(据/span>P.据/span>1据/span>-据/span>问:据/span>1据/span>)据/span>=据/span>0.据/span>那据/span>
暗示据span class="katex">
问:据/span>2据/span>=据/span>0.据/span>=据/span>P.据/span>2据/span>或者据span class="katex">
P.据/span>1据/span>=据/span>问:据/span>1据/span>。因此我们需要两者据span class="katex">
P.据/span>和据span class="katex">
问:据/span>真实的,或据span class="katex">
P.据/span>和据span class="katex">
问:据/span>是彼此的复共轭。据/P.>据B.lockquote class="example">
找到一个二次的根源据span class="katex">
2据/span>和据span class="katex">
5.据/span>。据/P.>据hr>
让二次BE据span class="katex">
X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>,在这里我们希望找到据span class="katex">
B.据/span>和据span class="katex">
C据/span>。然后vieta的配方告诉我们据/P.>据P.>据span class="katex-display">
B.据/span>=据/span>-据/span>(据/span>2据/span>+据/span>5.据/span>)据/span>=据/span>-据/span>7.据/span>那据/span>C据/span>=据/span>2据/span>×据/span>5.据/span>=据/span>1据/span>0.据/span>。据/span>
因此所期望的是二次据span class="katex">
X据/span>2据/span>-据/span>7.据/span>X据/span>+据/span>1据/span>0.据/span>。据/span>
□据/span>
找到一个二次的根源据span class="katex">
3.据/span>+据/span>2据/span>一世据/span>和据span class="katex">
3.据/span>-据/span>2据/span>一世据/span>。据/P.>据hr>
我们所期望的二次BE据span class="katex">
X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>,在这里我们希望找到据span class="katex">
B.据/span>和据span class="katex">
C据/span>。然后韦达的公式告诉我们据/P.>据P.>据span class="katex-display">
B.据/span>=据/span>-据/span>[据/span>(据/span>3.据/span>+据/span>2据/span>一世据/span>)据/span>+据/span>(据/span>3.据/span>-据/span>2据/span>一世据/span>)据/span>]据/span>=据/span>-据/span>6.据/span>那据/span>C据/span>=据/span>(据/span>3.据/span>+据/span>2据/span>一世据/span>)据/span>(据/span>3.据/span>-据/span>2据/span>一世据/span>)据/span>=据/span>1据/span>3.据/span>。据/span>
所以所需的二次是据span class="katex">
X据/span>2据/span>-据/span>6.据/span>X据/span>+据/span>1据/span>3.据/span>。据/span>
□据/span>
笔记据/strong>:由于根部是彼此复杂的缀合物,所以二次的系数变得真实。据/P.>据!-- end-example -->
解决方程系统据/P.>据P.>据span class="katex-display">
一种据/span>+据/span>B.据/span>=据/span>7.据/span>那据/span>一种据/span>B.据/span>=据/span>1据/span>0.据/span>。据/span>
由韦达的公式,我们知道,据span class="katex">
一种据/span>和据span class="katex">
B.据/span>是方程的根据span class="katex">
X据/span>2据/span>-据/span>7.据/span>X据/span>+据/span>1据/span>0.据/span>=据/span>0.据/span>。既然我们可以因式分解它作为据span class="katex">
(据/span>X据/span>-据/span>2据/span>)据/span>(据/span>X据/span>-据/span>5.据/span>)据/span>=据/span>0.据/span>,我们得到了据span class="katex">
{据/span>一种据/span>那据/span>B.据/span>}据/span>=据/span>{据/span>2据/span>那据/span>5.据/span>}据/span>。据/span>
□据/span>
泛化到更高的多项式据/h2>
考虑具有复杂系数和根的二次方程据span class="katex">
R.据/span>1据/span>那据span class="katex">
R.据/span>2据/span>:据/span>
一种据/span>2据/span>X据/span>2据/span>+据/span>一种据/span>1据/span>X据/span>+据/span>一种据/span>0.据/span>=据/span>一种据/span>2据/span>(据/span>X据/span>-据/span>R.据/span>1据/span>)据/span>(据/span>X据/span>-据/span>R.据/span>2据/span>)据/span>。据/span>
通过比较系数,我们可以看到,据/P.>据P.>据span class="katex-display">
R.据/span>1据/span>+据/span>R.据/span>2据/span>=据/span>-据/span>一种据/span>2据/span>一种据/span>1据/span>那据/span>R.据/span>1据/span>R.据/span>2据/span>=据/span>一种据/span>2据/span>一种据/span>0.据/span>。据/span>
这给出了多项式的根系与多项式的系数之间的关系。概括了学位多项式的想法据span class="katex">
N.据/span>,我们有以下的公式:据/P.>据B.lockquote class="definition">
vieta的惯例据/strong>:据/P.>据P.>让据span class="katex">
P.据/span>(据/span>X据/span>)据/span>=据/span>一种据/span>N.据/span>X据/span>N.据/span>+据/span>一种据/span>N.据/span>-据/span>1据/span>X据/span>N.据/span>-据/span>1据/span>+据/span>⋯据/span>+据/span>一种据/span>0.据/span>与复杂系数的多项式据span class="katex">
N.据/span>,复杂的根据span class="katex">
R.据/span>N.据/span>那据/span>R.据/span>N.据/span>-据/span>1据/span>那据/span>......据/span>那据/span>R.据/span>1据/span>。那么对于任何整数据span class="katex">
0.据/span>≤.据/span>K.据/span>≤.据/span>N.据/span>那据/span>
1据/span>≤.据/span>一世据/span>1据/span>据据/span>一世据/span>2据/span>据据/span>⋯据/span>据据/span>一世据/span>K.据/span>≤.据/span>N.据/span>σ.据/span>R.据/span>一世据/span>1据/span>R.据/span>一世据/span>2据/span>⋯据/span>R.据/span>一世据/span>K.据/span>=据/span>(据/span>-据/span>1据/span>)据/span>K.据/span>一种据/span>N.据/span>一种据/span>N.据/span>-据/span>K.据/span>。据/span>□据/span>
Vieta公式左侧的表达是基本的对称功能据span class="katex">
R.据/span>1据/span>那据/span>R.据/span>2据/span>那据/span>......据/span>R.据/span>N.据/span>。韦达公式的证明遵循公式中的系数比较据/P.>据P.>据span class="katex-display">
一种据/span>N.据/span>X据/span>N.据/span>+据/span>一种据/span>N.据/span>-据/span>1据/span>X据/span>N.据/span>-据/span>1据/span>+据/span>⋯据/span>+据/span>一种据/span>0.据/span>=据/span>一种据/span>N.据/span>(据/span>X据/span>-据/span>R.据/span>1据/span>)据/span>(据/span>X据/span>-据/span>R.据/span>2据/span>)据/span>(据/span>X据/span>-据/span>R.据/span>3.据/span>)据/span>⋯据/span>(据/span>X据/span>-据/span>R.据/span>N.据/span>)据/span>。据/span>
作为二次情况下,韦达的公式给出了一个公式来找到根的总和:据/P.>据P.>据span class="katex-display">
一世据/span>=据/span>1据/span>σ.据/span>N.据/span>R.据/span>一世据/span>=据/span>-据/span>一种据/span>N.据/span>一种据/span>N.据/span>-据/span>1据/span>。据/span>
同样,我们有以下产品的以下等式:据/P.>据P.>据span class="katex-display">
R.据/span>1据/span>R.据/span>2据/span>⋯据/span>R.据/span>N.据/span>=据/span>(据/span>-据/span>1据/span>)据/span>N.据/span>一种据/span>N.据/span>一种据/span>0.据/span>。据/span>
Vieta的公式为多项式根系和系数之间的关系提供了往往在解决问题的过程中的关系。据/P.>据B.lockquote class="example">
认为据span class="katex">
K.据/span>是一个数字,使得三次多项式据span class="katex">
P.据/span>(据/span>X据/span>)据/span>=据/span>-据/span>2据/span>X据/span>3.据/span>+据/span>4.据/span>8.据/span>X据/span>2据/span>+据/span>K.据/span>有三根整数根本是所有素数。有多少可能的不同价值据span class="katex">
K.据/span>还是据/span>
让据span class="katex">
P.据/span>那据/span>问:据/span>那据/span>和据span class="katex">
R.据/span>表示三个整数根据span class="katex">
P.据/span>(据/span>X据/span>)据/span>。然后由Vieta的公式,我们有据span class="katex">
P.据/span>问:据/span>+据/span>问:据/span>R.据/span>+据/span>P.据/span>R.据/span>=据/span>0.据/span>, 但由于据span class="katex">
P.据/span>那据/span>问:据/span>和据span class="katex">
R.据/span>是素数,他们每个人都严格大于1,因此没有这样的据span class="katex">
P.据/span>(据/span>X据/span>)据/span>存在。据span class="katex">
□据/span>
[1968普特南考试]找到所有多项式据span class="katex">
P.据/span>(据/span>X据/span>)据/span>=据/span>X据/span>N.据/span>+据/span>一种据/span>N.据/span>-据/span>1据/span>X据/span>N.据/span>-据/span>1据/span>+据/span>⋯据/span>+据/span>一种据/span>0.据/span>这样据span class="katex">
一种据/span>一世据/span>=据/span>±据/span>1据/span>对所有人据span class="katex">
0.据/span>≤.据/span>一世据/span>≤.据/span>N.据/span>-据/span>1据/span>满足所有根的条件据span class="katex">
P.据/span>(据/span>X据/span>)据/span>是真实的。据/P.>据hr>
如果据span class="katex">
R.据/span>1据/span>那据/span>R.据/span>2据/span>那据/span>⋯据/span>那据/span>R.据/span>N.据/span>是真正的根源据span class="katex">
P.据/span>(据/span>X据/span>)据/span>,然后韦达公式暗示据/P.>据P.>据span class="katex-display">
一世据/span>=据/span>1据/span>σ.据/span>N.据/span>R.据/span>一世据/span>=据/span>-据/span>一种据/span>N.据/span>-据/span>1据/span>和据/span>1据/span>≤.据/span>一世据/span>据据/span>j据/span>≤.据/span>N.据/span>σ.据/span>R.据/span>一世据/span>R.据/span>j据/span>=据/span>一种据/span>N.据/span>-据/span>2据/span>那据/span>
这意味着据/P.>据P.>据span class="katex-display">
一世据/span>=据/span>1据/span>σ.据/span>N.据/span>R.据/span>一世据/span>2据/span>=据/span>(据/span>一世据/span>=据/span>1据/span>σ.据/span>N.据/span>R.据/span>一世据/span>)据/span>2据/span>-据/span>2据/span>(据/span>1据/span>≤.据/span>一世据/span>据据/span>j据/span>≤.据/span>N.据/span>σ.据/span>R.据/span>一世据/span>R.据/span>j据/span>)据/span>=据/span>一种据/span>N.据/span>-据/span>1据/span>2据/span>-据/span>2据/span>一种据/span>N.据/span>-据/span>2据/span>≤.据/span>3.据/span>。据/span>
请注意,这也表明,据span class="katex">
一种据/span>N.据/span>-据/span>1据/span>2据/span>-据/span>2据/span>一种据/span>N.据/span>-据/span>2据/span>≥据/span>0.据/span>因为左边是实数的平方和。自从据span class="katex">
一种据/span>N.据/span>-据/span>1据/span>=据/span>±据/span>1据/span>, 这意味着据span class="katex">
一种据/span>N.据/span>-据/span>2据/span>=据/span>-据/span>1据/span>。此外,通过韦达的公式,据span class="katex">
(据/span>R.据/span>1据/span>R.据/span>2据/span>⋯据/span>R.据/span>N.据/span>)据/span>2据/span>=据/span>1据/span>。然后由据一种target="_blank" rel="nofollow" href="//www.parkandroid.com/wiki/arithmetic-mean-geometric-mean/">AM-GM不等式据/一种>, 我们有据span class="katex">
σ.据/span>一世据/span>=据/span>1据/span>N.据/span>R.据/span>一世据/span>2据/span>≥据/span>N.据/span>,这意味着据span class="katex">
N.据/span>≤.据/span>3.据/span>。然后我们可以枚举所有此类多项式以找到解决方案据span class="katex">
X据/span>±据/span>1据/span>那据span class="katex">
X据/span>2据/span>±据/span>X据/span>-据/span>1据/span>, 和据span class="katex">
X据/span>3.据/span>-据/span>X据/span>±据/span>(据/span>X据/span>2据/span>-据/span>1据/span>)据/span>。据span class="katex">
□据/span>
让据span class="katex">
R.据/span>1据/span>那据/span>R.据/span>2据/span>和据span class="katex">
R.据/span>3.据/span>是多项式的根据span class="katex">
5.据/span>X据/span>3.据/span>-据/span>1据/span>1据/span>X据/span>2据/span>+据/span>7.据/span>X据/span>+据/span>3.据/span>。评估据span class="katex">
R.据/span>1据/span>3.据/span>+据/span>R.据/span>2据/span>3.据/span>+据/span>R.据/span>3.据/span>3.据/span>。据/span>
在这个问题上,韦达定理的应用是不是很明显,而且表达必须改变。从据一种target="_blank" rel="nofollow" href="//www.parkandroid.com/assessment/techniques-trainer/advanced-factorization/">因式分解4据/一种>, 我们有据/P.>据P.>据span class="katex-display">
R.据/span>1据/span>3.据/span>+据/span>R.据/span>2据/span>3.据/span>+据/span>R.据/span>3.据/span>3.据/span>=据/span>3.据/span>R.据/span>1据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>[据/span>R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>-据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>]据/span>。据/span>
现在,我们只需要知道如何计算据span class="katex">
R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>。再次,从分解,我们有据/P.>据P.>据span class="katex-display">
R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>=据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>2据/span>-据/span>2据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>那据/span>
这使得我们可以得出这样的结论据/P.>据P.>据span class="katex-display">
R.据/span>1据/span>3.据/span>+据/span>R.据/span>2据/span>3.据/span>+据/span>R.据/span>3.据/span>3.据/span>=据/span>3.据/span>R.据/span>1据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>[据/span>R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>-据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>]据/span>=据/span>3.据/span>R.据/span>1据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>[据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>2据/span>-据/span>3.据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>]据/span>=据/span>3.据/span>×据/span>(据/span>-据/span>5.据/span>3.据/span>)据/span>+据/span>(据/span>5.据/span>1据/span>1据/span>)据/span>[据/span>(据/span>5.据/span>1据/span>1据/span>)据/span>2据/span>-据/span>3.据/span>×据/span>5.据/span>7.据/span>]据/span>=据/span>-据/span>1据/span>2据/span>5.据/span>4.据/span>9.据/span>。据/span>□据/span>
韦达的公式解决问题 - 基本据/h2>
让据span class="katex">
一种据/span>和据span class="katex">
B.据/span>是2张不同的实数,且据span class="katex">
一种据/span>2据/span>+据/span>3.据/span>一种据/span>+据/span>1据/span>=据/span>B.据/span>2据/span>+据/span>3.据/span>B.据/span>+据/span>1据/span>=据/span>0.据/span>。找到价值据span class="katex">
B.据/span>一种据/span>+据/span>一种据/span>B.据/span>。据/P.>据/div>
查找满足以下方程式系统的所有三倍的复数:据/P.>据P.>据span class="katex-display">
一种据/span>+据/span>B.据/span>+据/span>C据/span>一种据/span>B.据/span>+据/span>B.据/span>C据/span>+据/span>C据/span>一种据/span>一种据/span>B.据/span>C据/span>=据/span>0.据/span>=据/span>0.据/span>=据/span>0.据/span>。据/span>
任何据span class="katex">
3.据/span>数字据span class="katex">
(据/span>一种据/span>那据/span>B.据/span>那据/span>C据/span>)据/span>可以被认为是所述首一三次多项式的根据/P.>据P.>据span class="katex-display">
P.据/span>(据/span>X据/span>)据/span>=据/span>(据/span>X据/span>-据/span>一种据/span>)据/span>(据/span>X据/span>-据/span>B.据/span>)据/span>(据/span>X据/span>-据/span>C据/span>)据/span>。据/span>
应用韦达的公式多项式,我们得到据span class="katex">
P.据/span>(据/span>X据/span>)据/span>=据/span>X据/span>3.据/span>。因此,唯一的根据span class="katex">
P.据/span>(据/span>X据/span>)据/span>是据span class="katex">
X据/span>=据/span>0.据/span>(与多重据span class="katex">
3.据/span>),这意味着据span class="katex">
(据/span>0.据/span>那据/span>0.据/span>那据/span>0.据/span>)据/span>是唯一的三重复数的一种满足式的系统。据span class="katex">
□据/span>
让据span class="katex">
R.据/span>1据/span>那据/span>R.据/span>2据/span>和据span class="katex">
R.据/span>3.据/span>是多项式的根据span class="katex">
5.据/span>X据/span>3.据/span>-据/span>1据/span>1据/span>X据/span>2据/span>+据/span>7.据/span>X据/span>+据/span>3.据/span>。评估据/P.>据P.>据span class="katex-display">
R.据/span>1据/span>(据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>+据/span>R.据/span>2据/span>(据/span>1据/span>+据/span>R.据/span>3.据/span>+据/span>R.据/span>1据/span>)据/span>+据/span>R.据/span>3.据/span>(据/span>1据/span>+据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>)据/span>。据/span>
表达式等于据span class="katex">
R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>+据/span>2据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>。由韦达的公式,我们知道,据/P.>据P.>据span class="katex-display">
R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>=据/span>-据/span>5.据/span>-据/span>1据/span>1据/span>=据/span>5.据/span>1据/span>1据/span>那据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>=据/span>5.据/span>7.据/span>那据/span>R.据/span>1据/span>R.据/span>2据/span>R.据/span>3.据/span>=据/span>-据/span>5.据/span>3.据/span>那据/span>
所以据/P.>据P.>据span class="katex-display">
R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>+据/span>2据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>=据/span>5.据/span>1据/span>1据/span>+据/span>2据/span>×据/span>5.据/span>7.据/span>=据/span>5.据/span>2据/span>5.据/span>=据/span>5.据/span>。据/span>□据/span>
笔记据/strong>:一种常见的方法是尝试找到多项式的每个根源,特别是因为我们知道一个根源必须是真实的(为什么?)。但是,这不一定是一个可行的选择,因为我们很难确定实际的根源。据/em>
vieta的配方问题解决 - 中间据/h2>
让据span class="katex">
R.据/span>1据/span>那据/span>R.据/span>2据/span>和据span class="katex">
R.据/span>3.据/span>是多项式的根据span class="katex">
5.据/span>X据/span>3.据/span>-据/span>1据/span>1据/span>X据/span>2据/span>+据/span>7.据/span>X据/span>+据/span>3.据/span>。评估据span class="katex">
R.据/span>1据/span>3.据/span>+据/span>R.据/span>2据/span>3.据/span>+据/span>R.据/span>3.据/span>3.据/span>。据/span>
在这个问题上,韦达定理的应用是不是很明显,而且表达必须改变。从分解,我们有据/P.>据P.>据span class="katex-display">
R.据/span>1据/span>3.据/span>+据/span>R.据/span>2据/span>3.据/span>+据/span>R.据/span>3.据/span>3.据/span>=据/span>3.据/span>R.据/span>1据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>[据/span>R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>-据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>]据/span>。据/span>
现在,我们只需要知道如何计算据span class="katex">
R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>。再次,从分解,我们有据/P.>据P.>据span class="katex-display">
R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>=据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>2据/span>-据/span>2据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>那据/span>
这使得我们可以得出这样的结论据/P.>据P.>据span class="katex-display">
R.据/span>1据/span>3.据/span>+据/span>R.据/span>2据/span>3.据/span>+据/span>R.据/span>3.据/span>3.据/span>=据/span>3.据/span>R.据/span>1据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>[据/span>R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>-据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>]据/span>=据/span>3.据/span>R.据/span>1据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>[据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>2据/span>-据/span>3.据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>]据/span>=据/span>3.据/span>×据/span>(据/span>-据/span>5.据/span>3.据/span>)据/span>+据/span>(据/span>5.据/span>1据/span>1据/span>)据/span>[据/span>(据/span>5.据/span>1据/span>1据/span>)据/span>2据/span>-据/span>3.据/span>×据/span>5.据/span>7.据/span>]据/span>=据/span>-据/span>1据/span>2据/span>5.据/span>4.据/span>9.据/span>。据/span>□据/span>
如果据span class="katex">
α.据/span>那据/span>β据/span>是根茎方程据span class="katex">
X据/span>2据/span>+据/span>2据/span>X据/span>+据/span>3.据/span>=据/span>0.据/span>,什么是二次方程式,其根据span class="katex">
(据/span>α.据/span>-据/span>α.据/span>1据/span>)据/span>2据/span>和据span class="katex">
(据/span>β据/span>-据/span>β据/span>1据/span>)据/span>2据/span>还是据/span>
[arml 2012,团队问题,#6]据/P.>据P.>的零据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>X据/span>6.据/span>+据/span>2据/span>X据/span>5.据/span>+据/span>3.据/span>X据/span>4.据/span>+据/span>5.据/span>X据/span>3.据/span>+据/span>8.据/span>X据/span>2据/span>+据/span>1据/span>3.据/span>X据/span>+据/span>2据/span>1据/span>是不同的复数。计算的平均值据span class="katex">
一种据/span>+据/span>B.据/span>C据/span>+据/span>D.据/span>E.据/span>F据/span>超过所有可能的排列据span class="katex">
(据/span>一种据/span>那据/span>B.据/span>那据/span>C据/span>那据/span>D.据/span>那据/span>E.据/span>那据/span>F据/span>)据/span>这些六个号码。据/P.>据hr>
韦达的公式解决问题 - 高级据/h2>
[IMO名单]据B.R.>确定参数的所有实值据span class="katex">
一种据/span>为此公式据/P.>据P.>据span class="katex-display">
1据/span>6.据/span>X据/span>4.据/span>-据/span>一种据/span>X据/span>3.据/span>+据/span>(据/span>2据/span>一种据/span>+据/span>1据/span>7.据/span>)据/span>X据/span>2据/span>-据/span>一种据/span>X据/span>+据/span>1据/span>6.据/span>=据/span>0.据/span>
具有形成一几何级数恰好四个不同的实根。据/P.>据hr>
假设据span class="katex">
一种据/span>满足了这个问题,这要求据span class="katex">
X据/span>那据span class="katex">
问:据/span>X据/span>那据span class="katex">
问:据/span>2据/span>X据/span>那据span class="katex">
问:据/span>3.据/span>X据/span>是给定方程的根。然后据span class="katex">
X据/span>据/span>=据/span>0.据/span>我们可以假设据span class="katex">
|据/span>问:据/span>|据/span>>据/span>1据/span>, 以便据span class="katex">
|据/span>X据/span>|据/span>据据/span>|据/span>问:据/span>X据/span>|据/span>据据/span>|据/span>问:据/span>2据/span>X据/span>|据/span>据据/span>|据/span>问:据/span>3.据/span>X据/span>|据/span>。请注意,该系数是对称的,即,第一系数是相同的第五个,第二个是相同的第四,三是一样的三分之一。它保证了我们,如果据span class="katex">
α.据/span>是根,则其倒数(这是据span class="katex">
α.据/span>-据/span>1据/span>=据/span>α.据/span>1据/span>)也将是一个根目录。因此,据span class="katex">
X据/span>1据/span>=据/span>问:据/span>3.据/span>X据/span>那据/span>所以据span class="katex">
问:据/span>=据/span>X据/span>-据/span>3.据/span>2据/span>和根据span class="katex">
X据/span>那据/span>X据/span>3.据/span>1据/span>那据/span>X据/span>3.据/span>-据/span>1据/span>那据/span>X据/span>-据/span>1据/span>。据/P.>据P.>现在,由韦达的公式,我们有据span class="katex">
X据/span>+据/span>X据/span>3.据/span>1据/span>+据/span>X据/span>3.据/span>-据/span>1据/span>+据/span>X据/span>-据/span>1据/span>=据/span>1据/span>6.据/span>一种据/span>和据span class="katex">
X据/span>3.据/span>4.据/span>+据/span>X据/span>3.据/span>2据/span>+据/span>1据/span>+据/span>1据/span>+据/span>X据/span>3.据/span>-据/span>2据/span>+据/span>X据/span>3.据/span>-据/span>4.据/span>=据/span>1据/span>6.据/span>2据/span>一种据/span>+据/span>1据/span>7.据/span>。在环境据span class="katex">
Z.据/span>=据/span>X据/span>3.据/span>1据/span>+据/span>X据/span>3.据/span>-据/span>1据/span>这些方程成为据/P.>据P.>据span class="katex-display">
Z.据/span>3.据/span>-据/span>2据/span>Z.据/span>=据/span>1据/span>6.据/span>一种据/span>那据/span>(据/span>Z.据/span>2据/span>-据/span>2据/span>)据/span>2据/span>+据/span>Z.据/span>2据/span>-据/span>2据/span>=据/span>1据/span>6.据/span>2据/span>一种据/span>+据/span>1据/span>7.据/span>。据/span>
替代据span class="katex">
一种据/span>=据/span>1据/span>6.据/span>(据/span>Z.据/span>3.据/span>-据/span>2据/span>Z.据/span>)据/span>在第二个公式导致据span class="katex">
Z.据/span>4.据/span>-据/span>2据/span>Z.据/span>3.据/span>-据/span>3.据/span>Z.据/span>2据/span>+据/span>4.据/span>Z.据/span>+据/span>1据/span>6.据/span>1据/span>5.据/span>=据/span>0.据/span>。我们观察到,该多项式因素据span class="katex">
(据/span>Z.据/span>+据/span>2据/span>3.据/span>)据/span>(据/span>Z.据/span>-据/span>2据/span>5.据/span>)据/span>(据/span>Z.据/span>2据/span>-据/span>Z.据/span>-据/span>4.据/span>1据/span>)据/span>。自从据span class="katex">
|据/span>Z.据/span>|据/span>=据/span>|据/span>|据/span>|据/span>X据/span>3.据/span>1据/span>+据/span>X据/span>-据/span>3.据/span>1据/span>|据/span>|据/span>|据/span>≥据/span>2据/span>,唯一可行的价值是据span class="katex">
Z.据/span>=据/span>2据/span>5.据/span>。最后据span class="katex">
一种据/span>=据/span>1据/span>7.据/span>0.据/span>。据/span>
整理上式,我们得到据span class="katex">
1据/span>6.据/span>(据/span>X据/span>2据/span>+据/span>X据/span>2据/span>1据/span>)据/span>-据/span>1据/span>7.据/span>0.据/span>(据/span>X据/span>+据/span>X据/span>1据/span>)据/span>+据/span>3.据/span>5.据/span>7.据/span>=据/span>0.据/span>。为了简化计算,我们可以调用据span class="katex">
y据/span>=据/span>X据/span>+据/span>X据/span>1据/span>因此据span class="katex">
y据/span>2据/span>=据/span>X据/span>2据/span>+据/span>2据/span>+据/span>X据/span>2据/span>1据/span>,从而获得一种新形式据/P.>据P.>据span class="katex-display">
1据/span>6.据/span>(据/span>y据/span>2据/span>-据/span>2据/span>)据/span>-据/span>1据/span>7.据/span>0.据/span>y据/span>+据/span>3.据/span>5.据/span>7.据/span>=据/span>0.据/span>⟹据/span>1据/span>6.据/span>y据/span>2据/span>-据/span>1据/span>7.据/span>0.据/span>y据/span>+据/span>3.据/span>2据/span>5.据/span>=据/span>0.据/span>那据/span>
其根源是据span class="katex">
2据/span>5.据/span>和据span class="katex">
8.据/span>6.据/span>5.据/span>。我们必须插上两回据span class="katex">
y据/span>=据/span>X据/span>+据/span>X据/span>1据/span>,导致我们两个二次方程式,最后得到据span class="katex-display">
8.据/span>1据/span>那据/span>2据/span>1据/span>那据/span>2据/span>那据/span>8.据/span>。据/span>□据/span>
(由...制作据一种target="_blank" rel="nofollow" href="//www.parkandroid.com/profile/mark-4vl4ha/feed/">标记Hennings.据/一种>)据B.R.>显示据/P.>据P.>据span class="katex-display">
j据/span>=据/span>1据/span>σ.据/span>N.据/span>婴儿床据/span>2据/span>(据/span>2据/span>N.据/span>+据/span>1据/span>j据/span>π据/span>)据/span>=据/span>3.据/span>1据/span>N.据/span>(据/span>2据/span>N.据/span>-据/span>1据/span>)据/span>
对于任何整数据span class="katex">
N.据/span>≥据/span>1据/span>。据/P.>据hr>
(解决方案据一种target="_blank" rel="nofollow" href="//www.parkandroid.com/profile/aditya-85fq8c/">阿迪亚夏尔马据/一种>)据/P.>据P.>让我们先从德棣美弗定理任意正整数据span class="katex">
m据/span>:据/span>
(据/span>COS.据/span>X据/span>+据/span>一世据/span>罪据/span>X据/span>)据/span>m据/span>罪据/span>m据/span>X据/span>COS.据/span>m据/span>X据/span>+据/span>一世据/span>罪据/span>m据/span>X据/span>=据/span>COS.据/span>m据/span>X据/span>+据/span>一世据/span>罪据/span>m据/span>X据/span>=据/span>(据/span>婴儿床据/span>X据/span>+据/span>一世据/span>)据/span>m据/span>=据/span>婴儿床据/span>m据/span>X据/span>+据/span>(据/span>1据/span>m据/span>)据/span>婴儿床据/span>m据/span>-据/span>1据/span>(据/span>X据/span>)据/span>一世据/span>+据/span>⋯据/span>+据/span>(据/span>m据/span>-据/span>1据/span>m据/span>)据/span>婴儿床据/span>(据/span>X据/span>)据/span>一世据/span>m据/span>-据/span>1据/span>+据/span>一世据/span>m据/span>。据/span>
我们可以组RHS如下,因为我们有据span class="katex">
一世据/span>2据/span>=据/span>-据/span>1据/span>:据/span>
RHS.据/span>=据/span>(据/span>婴儿床据/span>m据/span>X据/span>-据/span>(据/span>2据/span>m据/span>)据/span>婴儿床据/span>m据/span>-据/span>2据/span>X据/span>+据/span>⋯据/span>)据/span>+据/span>一世据/span>(据/span>(据/span>1据/span>m据/span>)据/span>婴儿床据/span>m据/span>-据/span>1据/span>X据/span>-据/span>(据/span>3.据/span>m据/span>)据/span>婴儿床据/span>m据/span>-据/span>3.据/span>X据/span>+据/span>⋯据/span>)据/span>。据/span>
等同于LHS和RHS虚部,我们得到据/P.>据P.>据span class="katex-display">
(据/span>1据/span>m据/span>)据/span>婴儿床据/span>m据/span>-据/span>1据/span>X据/span>-据/span>(据/span>3.据/span>m据/span>)据/span>婴儿床据/span>m据/span>-据/span>3.据/span>X据/span>+据/span>⋯据/span>=据/span>罪据/span>m据/span>X据/span>罪据/span>m据/span>X据/span>。据/span>
现在,如果我们让据span class="katex">
m据/span>=据/span>2据/span>N.据/span>+据/span>1据/span>和据span class="katex">
X据/span>=据/span>m据/span>j据/span>π据/span>对于任何一个据span class="katex">
j据/span>=据/span>1据/span>那据/span>2据/span>那据/span>......据/span>那据/span>m据/span>,则方程简化为据/P.>据P.>据span class="katex-display">
(据/span>1据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>婴儿床据/span>2据/span>N.据/span>X据/span>-据/span>(据/span>3.据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>婴儿床据/span>2据/span>N.据/span>-据/span>2据/span>X据/span>+据/span>⋯据/span>=据/span>0.据/span>
自从据span class="katex">
罪据/span>m据/span>X据/span>=据/span>罪据/span>j据/span>π据/span>=据/span>0.据/span>。据/span>
让据span class="katex">
婴儿床据/span>2据/span>X据/span>=据/span>你据/span>那据/span>然后等式在据span class="katex">
你据/span>其根源之和由下式给出据span class="katex">
(据/span>1据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>(据/span>3.据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>因为我们知道从韦达的公式。自从据span class="katex">
X据/span>=据/span>2据/span>N.据/span>+据/span>1据/span>j据/span>π据/span>那据/span>
j据/span>=据/span>1据/span>σ.据/span>N.据/span>婴儿床据/span>2据/span>2据/span>N.据/span>+据/span>1据/span>j据/span>π据/span>=据/span>(据/span>1据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>(据/span>3.据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>=据/span>6.据/span>1据/span>2据/span>N.据/span>(据/span>2据/span>N.据/span>-据/span>1据/span>)据/span>=据/span>3.据/span>1据/span>N.据/span>(据/span>2据/span>N.据/span>-据/span>1据/span>)据/span>。据/span>□据/span>
什么是据/P.>据P.>据span class="katex-display">
K.据/span>=据/span>1据/span>σ.据/span>N.据/span>婴儿床据/span>4.据/span>(据/span>2据/span>N.据/span>+据/span>1据/span>K.据/span>π据/span>)据/span>还是据/span>
让据span class="katex">
W.据/span>是任何据一种href="//www.parkandroid.com/wiki/primitive-roots-of-unity/" class="wiki_link" title="原始" target="_blank">原始据/一种>据span class="katex">
(据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>TH.据/span>团结的根源。然后,据span class="katex">
P.据/span>(据/span>X据/span>)据/span>=据/span>X据/span>-据/span>1据/span>X据/span>2据/span>N.据/span>+据/span>1据/span>-据/span>1据/span>将有根源据span class="katex">
W.据/span>那据/span>W.据/span>2据/span>那据/span>......据/span>那据/span>W.据/span>2据/span>N.据/span>。我们也知道据span class="katex">
罪据/span>(据/span>2据/span>N.据/span>+据/span>1据/span>π据/span>K.据/span>)据/span>=据/span>2据/span>一世据/span>W.据/span>K.据/span>/据/span>2据/span>-据/span>W.据/span>-据/span>K.据/span>/据/span>2据/span>和据span class="katex">
COS.据/span>(据/span>2据/span>N.据/span>+据/span>1据/span>π据/span>K.据/span>)据/span>=据/span>2据/span>W.据/span>K.据/span>/据/span>2据/span>+据/span>W.据/span>-据/span>K.据/span>/据/span>2据/span>。然后据span class="katex">
婴儿床据/span>(据/span>2据/span>N.据/span>+据/span>1据/span>π据/span>K.据/span>)据/span>=据/span>W.据/span>K.据/span>-据/span>1据/span>一世据/span>(据/span>W.据/span>K.据/span>+据/span>1据/span>)据/span>。所以总和据span class="katex">
S.据/span>(说)是据/P.>据P.>据span class="katex-display">
S.据/span>=据/span>K.据/span>=据/span>1据/span>σ.据/span>N.据/span>(据/span>W.据/span>K.据/span>-据/span>1据/span>W.据/span>K.据/span>+据/span>1据/span>)据/span>4.据/span>。据/span>
现在,我们将通过做来做出根源的多项式转变据span class="katex">
y据/span>=据/span>X据/span>-据/span>1据/span>X据/span>+据/span>1据/span>, 然后据span class="katex">
X据/span>=据/span>y据/span>-据/span>1据/span>y据/span>+据/span>1据/span>。替换,在据span class="katex">
P.据/span>(据/span>X据/span>)据/span>:据/P.>据P.>据span class="katex-display">
P.据/span>(据/span>X据/span>)据/span>=据/span>y据/span>-据/span>1据/span>y据/span>+据/span>1据/span>-据/span>1据/span>(据/span>y据/span>-据/span>1据/span>y据/span>+据/span>1据/span>)据/span>2据/span>N.据/span>+据/span>1据/span>-据/span>1据/span>。据/span>
简化并使其成为一个多项式,我们得到据/P.>据P.>据span class="katex-display">
P.据/span>(据/span>y据/span>)据/span>=据/span>(据/span>y据/span>+据/span>1据/span>)据/span>2据/span>N.据/span>+据/span>1据/span>-据/span>(据/span>y据/span>-据/span>1据/span>)据/span>2据/span>N.据/span>+据/span>1据/span>。据/span>
扩展和再简化一下:据/P.>据P.>据span class="katex-display">
P.据/span>(据/span>y据/span>)据/span>=据/span>2据/span>(据/span>1据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>(据/span>y据/span>2据/span>)据/span>N.据/span>+据/span>2据/span>(据/span>3.据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>(据/span>y据/span>2据/span>)据/span>N.据/span>-据/span>1据/span>+据/span>2据/span>(据/span>5.据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>(据/span>y据/span>2据/span>)据/span>N.据/span>-据/span>2据/span>+据/span>⋯据/span>。据/span>
最后,由韦达定理,我们的总和据/P.>据P.>据span class="katex-display">
S.据/span>=据/span>K.据/span>=据/span>1据/span>σ.据/span>N.据/span>y据/span>K.据/span>4.据/span>=据/span>(据/span>K.据/span>=据/span>1据/span>σ.据/span>N.据/span>y据/span>K.据/span>2据/span>)据/span>2据/span>-据/span>2据/span>⎝据/span>⎛据/span>1据/span>≤.据/span>j据/span>据据/span>K.据/span>≤.据/span>N.据/span>σ.据/span>y据/span>j据/span>2据/span>y据/span>K.据/span>2据/span>⎠据/span>⎞据/span>=据/span>(据/span>2据/span>(据/span>1据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>-据/span>2据/span>(据/span>3.据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>)据/span>2据/span>-据/span>2据/span>(据/span>2据/span>(据/span>1据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>2据/span>(据/span>5.据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>)据/span>=据/span>9.据/span>N.据/span>2据/span>(据/span>2据/span>N.据/span>-据/span>1据/span>)据/span>2据/span>-据/span>1据/span>5.据/span>N.据/span>(据/span>2据/span>N.据/span>-据/span>1据/span>)据/span>(据/span>N.据/span>-据/span>1据/span>)据/span>(据/span>2据/span>N.据/span>-据/span>3.据/span>)据/span>=据/span>4.据/span>5.据/span>N.据/span>(据/span>2据/span>N.据/span>-据/span>1据/span>)据/span>(据/span>4.据/span>N.据/span>2据/span>+据/span>1据/span>0.据/span>N.据/span>-据/span>9.据/span>)据/span>。据/span>□据/span>
让据span class="katex">
X据/span>1据/span>那据/span>X据/span>2据/span>那据/span>。据/span>。据/span>。据/span>那据/span>X据/span>1据/span>0.据/span>是多项式的根据span class="katex">
X据/span>1据/span>0.据/span>+据/span>X据/span>9.据/span>+据/span>⋯据/span>+据/span>X据/span>+据/span>1据/span>。找到价值据span class="katex-display">
N.据/span>=据/span>1据/span>σ.据/span>1据/span>0.据/span>1据/span>-据/span>X据/span>N.据/span>1据/span>
韦达根跳据/h2>
主要文章:据一种href="//www.parkandroid.com/wiki/vieta-root-jumping/" class="wiki_link" title="韦达根跳" target="_blank">韦达根跳据/一种>据/P.>据!-- end-meta -->
韦达跳跃是一种特殊的下降法,已成为在更高层次的数学奥赛数论问题颇为流行的昵称。像血统的其他情况下,当你必须解决(方程,同余或不等式或系统),一个丢番图方程时,其解决方案有一定的据一种href="//www.parkandroid.com/wiki/subroutines/" class="wiki_link" title="递归" target="_blank">递归据/一种>结构体。据/P.>据/div>
也可以看看据/h2>
引用如下:据/strong>韦达的公式。据em>bright.org.据/em>。检索到据span class="retrieval-time">从据一种href="//www.parkandroid.com/wiki/vietas-formula/">//www.parkandroid.com/wiki/vietas-formula/据/一种>据/div>
硕士概念如此据/h4>
注册以阅读数学,科学和工程主题中的所有Wiki和测验。据/div>
×据/div>
问题加载......据/P.>据P.Class="note-text">注意加载......据/P.>据P.Class="set-text">设置加载......据/P.>据/div>
比赛数学二据/h3>
在AMC 10和12的水平下解决数学问题的指导训练。据/P.>据/div>
测验据/h4>
相关......据/h4>
- 代数据/span>>据/span>
韦达的公式据/strong>涉及的系数据一种href="//www.parkandroid.com/wiki/polynomials/" class="wiki_link" title="多项式" target="_blank">多项式据/一种>到的款项及其根源的产品,以及在群体采取根的产品。据/P.>据P.>例如,如果存在二次多项式据span class="katex"> F据/span>(据/span>X据/span>)据/span>=据/span>X据/span>2据/span>+据/span>2据/span>X据/span>-据/span>1据/span>5.据/span>,则其根据span class="katex"> X据/span>=据/span>-据/span>5.据/span>和据span class="katex"> X据/span>=据/span>3.据/span>, 因为据span class="katex"> F据/span>(据/span>X据/span>)据/span>=据/span>X据/span>2据/span>+据/span>2据/span>X据/span>-据/span>1据/span>5.据/span>=据/span>(据/span>X据/span>-据/span>3.据/span>)据/span>(据/span>X据/span>+据/span>5.据/span>)据/span>。韦达的公式可以找到根的总和据span class="katex"> (据/span>3.据/span>+据/span>(据/span>-据/span>5.据/span>)据/span>=据/span>-据/span>2据/span>)据/span>和根的产品据span class="katex"> (据/span>3.据/span>⋅据/span>(据/span>-据/span>5.据/span>)据/span>=据/span>-据/span>1据/span>5.据/span>)据/span>而不直接找到每一根。虽然这是在这个具体的例子相当琐碎,韦达的公式是在更复杂的代数多项式有许多根或当多项式的根不容易推导是非常有用的。对于一些问题,韦达的公式可以作为快捷方式寻求解决办法很快知道自己的根的款项或产品。据/P.>据/div>
内容据/h4>
韦达的公式 - 二次方程据/h2>
让我们从定义开始。据/P.>据B.lockquote class="definition">
韦达的公式二次方程式据/strong>:据/P.>据P.>给予据span class="katex"> F据/span>(据/span>X据/span>)据/span>=据/span>一种据/span>X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>如果方程据span class="katex"> F据/span>(据/span>X据/span>)据/span>=据/span>0.据/span>有根据span class="katex"> R.据/span>1据/span>和据span class="katex"> R.据/span>2据/span>, 然后据/P.>据P.>据span class="katex-display"> R.据/span>1据/span>+据/span>R.据/span>2据/span>=据/span>-据/span>一种据/span>B.据/span>那据/span>R.据/span>1据/span>R.据/span>2据/span>=据/span>一种据/span>C据/span>。据/span>□据/span>
这句话的证明在本节的最后给出。据/P.>据P.>我们立即看到的二次帮助系数如何确定根的关系。据/P.>据B.lockquote class="example">
如果据span class="katex">
α.据/span>和据span class="katex">
β据/span>是二次的根源据span class="katex">
X据/span>2据/span>-据/span>4.据/span>X据/span>+据/span>9.据/span>=据/span>0.据/span>什么是值据/P.>据ol>
从韦达的公式,我们认识到,据span class="katex">
α.据/span>+据/span>β据/span>=据/span>4.据/span>。据/P.>据/li>
从韦达的公式,我们认识到,据span class="katex">
α.据/span>β据/span>=据/span>9.据/span>。据/P.>据/li>
韦达的公式并没有告诉我们的价值据span class="katex">
α.据/span>2据/span>+据/span>β据/span>2据/span>直接地。我们需要做的,是写的据span class="katex">
α.据/span>2据/span>+据/span>β据/span>2据/span>按照据span class="katex">
α.据/span>+据/span>β据/span>和/或据span class="katex">
α.据/span>β据/span>,然后我们可以替换这些值。我们有据/P.>据/li>
α.据/span>2据/span>+据/span>β据/span>2据/span>=据/span>(据/span>α.据/span>+据/span>β据/span>)据/span>2据/span>-据/span>2据/span>α.据/span>β据/span>=据/span>4.据/span>2据/span>-据/span>2据/span>×据/span>9.据/span>=据/span>-据/span>2据/span>。据/span>□据/span> 笔记据/strong>:在这个问题上,根是复数据span class="katex">
2据/span>±据/span>5.据/span>
一世据/span>。如果我们要将这些产品作为价值观据span class="katex">
α.据/span>和据span class="katex">
β据/span>然后计算,我们必须使粗心犯错计算的机会较高。韦达的公式我们提供了一个更简单的方法。据/P.>据!-- end-example -->
你能快速猜出二次的根源吗?据span class="katex">
X据/span>2据/span>-据/span>5.据/span>X据/span>+据/span>6.据/span>还是据/span> 如果据span class="katex">
P.据/span>和据span class="katex">
问:据/span>是给定方程的根,然后韦达的公式告诉我们据/P.>据P.>据span class="katex-display">
P.据/span>+据/span>问:据/span>=据/span>5.据/span>那据/span>P.据/span>问:据/span>=据/span>6.据/span>那据/span> 这并不难看据span class="katex">
2据/span>+据/span>3.据/span>=据/span>5.据/span>和据span class="katex">
2据/span>×据/span>3.据/span>=据/span>6.据/span>。据/span>所以根必须是据span class="katex">
2据/span>和据span class="katex">
3.据/span>,他们确实是。据span class="katex">
□据/span> 给予据span class="katex">
α.据/span>和据span class="katex">
β据/span>是二次的根源据span class="katex">
一种据/span>X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>=据/span>0.据/span>, 表达据span class="katex">
一种据/span>2据/span>B.据/span>2据/span>-据/span>4.据/span>一种据/span>C据/span>按照据span class="katex">
α.据/span>和据span class="katex">
β据/span>。据/P.>据hr>
韦达的公式给了我们据span class="katex">
一种据/span>-据/span>B.据/span>=据/span>α.据/span>+据/span>β据/span>和据span class="katex">
一种据/span>C据/span>=据/span>α.据/span>β据/span>。取代这些,我们得到了据/P.>据P.>据span class="katex-display">
一种据/span>2据/span>B.据/span>2据/span>-据/span>4.据/span>一种据/span>C据/span>=据/span>一种据/span>2据/span>B.据/span>2据/span>-据/span>4.据/span>一种据/span>C据/span>=据/span>(据/span>α.据/span>+据/span>β据/span>)据/span>2据/span>-据/span>4.据/span>α.据/span>β据/span>=据/span>α.据/span>2据/span>+据/span>2据/span>α.据/span>β据/span>+据/span>β据/span>2据/span>-据/span>4.据/span>α.据/span>β据/span>=据/span>α.据/span>2据/span>-据/span>2据/span>α.据/span>β据/span>+据/span>β据/span>2据/span>=据/span>(据/span>α.据/span>-据/span>β据/span>)据/span>2据/span>。据/span>□据/span> 你可能认识据span class="katex">
B.据/span>2据/span>-据/span>4.据/span>一种据/span>C据/span>来自据一种href="//www.parkandroid.com/wiki/quadratic-formula/" class="wiki_link" title="二次公式" target="_blank">二次公式据/一种>。事实上,我们可以表明这一点据/P.>据P.>据span class="katex-display">
2据/span>一种据/span>-据/span>B.据/span>±据/span>B.据/span>2据/span>-据/span>4.据/span>一种据/span>C据/span>
=据/span>2据/span>1据/span>(据/span>一种据/span>-据/span>B.据/span>±据/span>一种据/span>2据/span>B.据/span>2据/span>-据/span>4.据/span>一种据/span>C据/span>
)据/span>=据/span>2据/span>(据/span>α.据/span>+据/span>β据/span>)据/span>±据/span>|据/span>α.据/span>-据/span>β据/span>|据/span>=据/span>α.据/span>或者据/span>β据/span>那据/span> 这证明了二次公式。据/P.>据B.lockquote class="proof">
由这件事据一种href="//www.parkandroid.com/wiki/remainder-factor-theorem/" class="wiki_link" title="剩余因子定理" target="_blank">剩余因子定理据/一种>由于多项式据span class="katex">
F据/span>(据/span>X据/span>)据/span>有根据span class="katex">
R.据/span>1据/span>和据span class="katex">
R.据/span>2据/span>,它必须有形式据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>一种据/span>(据/span>X据/span>-据/span>R.据/span>1据/span>)据/span>(据/span>X据/span>-据/span>R.据/span>2据/span>)据/span>=据/span>一种据/span>X据/span>2据/span>-据/span>一种据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>)据/span>X据/span>+据/span>一种据/span>R.据/span>1据/span>R.据/span>2据/span>对于某一常数据span class="katex">
一种据/span>。据/P.>据P.>比较系数据span class="katex">
F据/span>(据/span>X据/span>)据/span>=据/span>一种据/span>X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>,我们得出结论据span class="katex">
一种据/span>=据/span>一种据/span>那据/span>B.据/span>=据/span>-据/span>一种据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>)据/span>和据span class="katex">
C据/span>=据/span>一种据/span>R.据/span>1据/span>R.据/span>2据/span>。因此,我们得到据/P.>据P.>据span class="katex-display">
R.据/span>1据/span>+据/span>R.据/span>2据/span>=据/span>-据/span>一种据/span>B.据/span>那据/span>R.据/span>1据/span>R.据/span>2据/span>=据/span>一种据/span>C据/span>。据/span>□据/span>
韦达的公式 - 成型二次方程式据/h2>
让据span class="katex"> P.据/span>和据span class="katex"> 问:据/span>是首一元二次方程的真正根源据/P.>据P.>据span class="katex-display"> X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>=据/span>0.据/span>那据/span>(据/span>B.据/span>那据/span>C据/span>)据/span>∈据/span>R.据/span>2据/span>。据/span>
为什么摩尼?因为我们总是可以通过其领先的系数将整个方程来获得它的首一版本。所以,我们能说的据span class="katex"> B.据/span>那据/span>C据/span>还是自从据span class="katex"> P.据/span>和据span class="katex"> 问:据/span>此方程的根,我们可以因式分解公式为据/P.>据P.>据span class="katex-display"> X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>≡据/span>(据/span>X据/span>-据/span>P.据/span>)据/span>(据/span>X据/span>-据/span>问:据/span>)据/span>。据/span>
扩大右侧并重新排列,我们发现据/P.>据P.>据span class="katex-display"> X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>≡据/span>X据/span>2据/span>-据/span>(据/span>P.据/span>+据/span>问:据/span>)据/span>X据/span>+据/span>P.据/span>问:据/span>。据/span>
由于两个多项式相等,当且仅当它们的系数是相等的,由等同我们得到的系数据/P.>据P.>据span class="katex-display"> B.据/span>=据/span>-据/span>(据/span>P.据/span>+据/span>问:据/span>)据/span>那据/span>C据/span>=据/span>P.据/span>问:据/span>。据/span>
这就是所谓的韦达公式为二次多项式。它也同样能够扩展到多项式据一种target="_blank" rel="nofollow" href="#vietas-formula-higher-degrees">程度更高据/一种>。据/P.>据P.>根可以广泛地包括复数。也就是说,给定两个复数据span class="katex"> P.据/span>和据span class="katex"> 问:据/span>,我们总能构建一个首一二次,其根源是据span class="katex"> P.据/span>和据span class="katex"> 问:据/span>。更具体地讲,二次会据/P.>据P.>据span class="katex-display"> X据/span>2据/span>-据/span>(据/span>P.据/span>+据/span>问:据/span>)据/span>X据/span>+据/span>P.据/span>问:据/span>=据/span>0.据/span>。据/span>
这两个系数的时候会是真的吗?为了找到答案,集据span class="katex"> P.据/span>=据/span>P.据/span>1据/span>+据/span>P.据/span>2据/span>一世据/span>和据span class="katex"> 问:据/span>=据/span>问:据/span>1据/span>+据/span>问:据/span>2据/span>一世据/span>。然后系数是据/P.>据P.>据span class="katex-display"> B.据/span>=据/span>-据/span>(据/span>P.据/span>1据/span>+据/span>问:据/span>1据/span>)据/span>-据/span>(据/span>P.据/span>2据/span>+据/span>问:据/span>2据/span>)据/span>一世据/span>那据/span>C据/span>=据/span>(据/span>P.据/span>1据/span>问:据/span>1据/span>-据/span>P.据/span>2据/span>问:据/span>2据/span>)据/span>+据/span>(据/span>P.据/span>1据/span>问:据/span>2据/span>+据/span>P.据/span>2据/span>问:据/span>1据/span>)据/span>一世据/span>。据/span>
为了据span class="katex"> B.据/span>要真实,我们需要据span class="katex"> P.据/span>2据/span>+据/span>问:据/span>2据/span>=据/span>0.据/span>⇒据/span>P.据/span>2据/span>=据/span>-据/span>问:据/span>2据/span>。为了据span class="katex"> C据/span>要真实,我们需要据/P.>据P.>据span class="katex-display"> P.据/span>1据/span>问:据/span>2据/span>+据/span>P.据/span>2据/span>问:据/span>1据/span>=据/span>0.据/span>⇒据/span>问:据/span>2据/span>(据/span>P.据/span>1据/span>-据/span>问:据/span>1据/span>)据/span>=据/span>0.据/span>那据/span>
暗示据span class="katex">
问:据/span>2据/span>=据/span>0.据/span>=据/span>P.据/span>2据/span>或者据span class="katex">
P.据/span>1据/span>=据/span>问:据/span>1据/span>。因此我们需要两者据span class="katex">
P.据/span>和据span class="katex">
问:据/span>真实的,或据span class="katex">
P.据/span>和据span class="katex">
问:据/span>是彼此的复共轭。据/P.>据B.lockquote class="example">
找到一个二次的根源据span class="katex">
2据/span>和据span class="katex">
5.据/span>。据/P.>据hr>
让二次BE据span class="katex">
X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>,在这里我们希望找到据span class="katex">
B.据/span>和据span class="katex">
C据/span>。然后vieta的配方告诉我们据/P.>据P.>据span class="katex-display">
B.据/span>=据/span>-据/span>(据/span>2据/span>+据/span>5.据/span>)据/span>=据/span>-据/span>7.据/span>那据/span>C据/span>=据/span>2据/span>×据/span>5.据/span>=据/span>1据/span>0.据/span>。据/span> 因此所期望的是二次据span class="katex">
X据/span>2据/span>-据/span>7.据/span>X据/span>+据/span>1据/span>0.据/span>。据/span>
□据/span> 找到一个二次的根源据span class="katex">
3.据/span>+据/span>2据/span>一世据/span>和据span class="katex">
3.据/span>-据/span>2据/span>一世据/span>。据/P.>据hr>
我们所期望的二次BE据span class="katex">
X据/span>2据/span>+据/span>B.据/span>X据/span>+据/span>C据/span>,在这里我们希望找到据span class="katex">
B.据/span>和据span class="katex">
C据/span>。然后韦达的公式告诉我们据/P.>据P.>据span class="katex-display">
B.据/span>=据/span>-据/span>[据/span>(据/span>3.据/span>+据/span>2据/span>一世据/span>)据/span>+据/span>(据/span>3.据/span>-据/span>2据/span>一世据/span>)据/span>]据/span>=据/span>-据/span>6.据/span>那据/span>C据/span>=据/span>(据/span>3.据/span>+据/span>2据/span>一世据/span>)据/span>(据/span>3.据/span>-据/span>2据/span>一世据/span>)据/span>=据/span>1据/span>3.据/span>。据/span> 所以所需的二次是据span class="katex">
X据/span>2据/span>-据/span>6.据/span>X据/span>+据/span>1据/span>3.据/span>。据/span>
□据/span> 笔记据/strong>:由于根部是彼此复杂的缀合物,所以二次的系数变得真实。据/P.>据!-- end-example -->
解决方程系统据/P.>据P.>据span class="katex-display">
一种据/span>+据/span>B.据/span>=据/span>7.据/span>那据/span>一种据/span>B.据/span>=据/span>1据/span>0.据/span>。据/span> 由韦达的公式,我们知道,据span class="katex">
一种据/span>和据span class="katex">
B.据/span>是方程的根据span class="katex">
X据/span>2据/span>-据/span>7.据/span>X据/span>+据/span>1据/span>0.据/span>=据/span>0.据/span>。既然我们可以因式分解它作为据span class="katex">
(据/span>X据/span>-据/span>2据/span>)据/span>(据/span>X据/span>-据/span>5.据/span>)据/span>=据/span>0.据/span>,我们得到了据span class="katex">
{据/span>一种据/span>那据/span>B.据/span>}据/span>=据/span>{据/span>2据/span>那据/span>5.据/span>}据/span>。据/span>
□据/span>
泛化到更高的多项式据/h2>
考虑具有复杂系数和根的二次方程据span class="katex"> R.据/span>1据/span>那据span class="katex"> R.据/span>2据/span>:据/span>
一种据/span>2据/span>X据/span>2据/span>+据/span>一种据/span>1据/span>X据/span>+据/span>一种据/span>0.据/span>=据/span>一种据/span>2据/span>(据/span>X据/span>-据/span>R.据/span>1据/span>)据/span>(据/span>X据/span>-据/span>R.据/span>2据/span>)据/span>。据/span>
通过比较系数,我们可以看到,据/P.>据P.>据span class="katex-display"> R.据/span>1据/span>+据/span>R.据/span>2据/span>=据/span>-据/span>一种据/span>2据/span>一种据/span>1据/span>那据/span>R.据/span>1据/span>R.据/span>2据/span>=据/span>一种据/span>2据/span>一种据/span>0.据/span>。据/span>
这给出了多项式的根系与多项式的系数之间的关系。概括了学位多项式的想法据span class="katex"> N.据/span>,我们有以下的公式:据/P.>据B.lockquote class="definition">
vieta的惯例据/strong>:据/P.>据P.>让据span class="katex"> P.据/span>(据/span>X据/span>)据/span>=据/span>一种据/span>N.据/span>X据/span>N.据/span>+据/span>一种据/span>N.据/span>-据/span>1据/span>X据/span>N.据/span>-据/span>1据/span>+据/span>⋯据/span>+据/span>一种据/span>0.据/span>与复杂系数的多项式据span class="katex"> N.据/span>,复杂的根据span class="katex"> R.据/span>N.据/span>那据/span>R.据/span>N.据/span>-据/span>1据/span>那据/span>......据/span>那据/span>R.据/span>1据/span>。那么对于任何整数据span class="katex"> 0.据/span>≤.据/span>K.据/span>≤.据/span>N.据/span>那据/span>
1据/span>≤.据/span>一世据/span>1据/span>据据/span>一世据/span>2据/span>据据/span>⋯据/span>据据/span>一世据/span>K.据/span>≤.据/span>N.据/span>σ.据/span>R.据/span>一世据/span>1据/span>R.据/span>一世据/span>2据/span>⋯据/span>R.据/span>一世据/span>K.据/span>=据/span>(据/span>-据/span>1据/span>)据/span>K.据/span>一种据/span>N.据/span>一种据/span>N.据/span>-据/span>K.据/span>。据/span>□据/span>
Vieta公式左侧的表达是基本的对称功能据span class="katex"> R.据/span>1据/span>那据/span>R.据/span>2据/span>那据/span>......据/span>R.据/span>N.据/span>。韦达公式的证明遵循公式中的系数比较据/P.>据P.>据span class="katex-display"> 一种据/span>N.据/span>X据/span>N.据/span>+据/span>一种据/span>N.据/span>-据/span>1据/span>X据/span>N.据/span>-据/span>1据/span>+据/span>⋯据/span>+据/span>一种据/span>0.据/span>=据/span>一种据/span>N.据/span>(据/span>X据/span>-据/span>R.据/span>1据/span>)据/span>(据/span>X据/span>-据/span>R.据/span>2据/span>)据/span>(据/span>X据/span>-据/span>R.据/span>3.据/span>)据/span>⋯据/span>(据/span>X据/span>-据/span>R.据/span>N.据/span>)据/span>。据/span>
作为二次情况下,韦达的公式给出了一个公式来找到根的总和:据/P.>据P.>据span class="katex-display"> 一世据/span>=据/span>1据/span>σ.据/span>N.据/span>R.据/span>一世据/span>=据/span>-据/span>一种据/span>N.据/span>一种据/span>N.据/span>-据/span>1据/span>。据/span>
同样,我们有以下产品的以下等式:据/P.>据P.>据span class="katex-display"> R.据/span>1据/span>R.据/span>2据/span>⋯据/span>R.据/span>N.据/span>=据/span>(据/span>-据/span>1据/span>)据/span>N.据/span>一种据/span>N.据/span>一种据/span>0.据/span>。据/span>
Vieta的公式为多项式根系和系数之间的关系提供了往往在解决问题的过程中的关系。据/P.>据B.lockquote class="example">
认为据span class="katex"> K.据/span>是一个数字,使得三次多项式据span class="katex"> P.据/span>(据/span>X据/span>)据/span>=据/span>-据/span>2据/span>X据/span>3.据/span>+据/span>4.据/span>8.据/span>X据/span>2据/span>+据/span>K.据/span>有三根整数根本是所有素数。有多少可能的不同价值据span class="katex"> K.据/span>还是据/span>
让据span class="katex"> P.据/span>那据/span>问:据/span>那据/span>和据span class="katex"> R.据/span>表示三个整数根据span class="katex"> P.据/span>(据/span>X据/span>)据/span>。然后由Vieta的公式,我们有据span class="katex"> P.据/span>问:据/span>+据/span>问:据/span>R.据/span>+据/span>P.据/span>R.据/span>=据/span>0.据/span>, 但由于据span class="katex"> P.据/span>那据/span>问:据/span>和据span class="katex"> R.据/span>是素数,他们每个人都严格大于1,因此没有这样的据span class="katex"> P.据/span>(据/span>X据/span>)据/span>存在。据span class="katex"> □据/span>
[1968普特南考试]找到所有多项式据span class="katex"> P.据/span>(据/span>X据/span>)据/span>=据/span>X据/span>N.据/span>+据/span>一种据/span>N.据/span>-据/span>1据/span>X据/span>N.据/span>-据/span>1据/span>+据/span>⋯据/span>+据/span>一种据/span>0.据/span>这样据span class="katex"> 一种据/span>一世据/span>=据/span>±据/span>1据/span>对所有人据span class="katex"> 0.据/span>≤.据/span>一世据/span>≤.据/span>N.据/span>-据/span>1据/span>满足所有根的条件据span class="katex"> P.据/span>(据/span>X据/span>)据/span>是真实的。据/P.>据hr>
如果据span class="katex"> R.据/span>1据/span>那据/span>R.据/span>2据/span>那据/span>⋯据/span>那据/span>R.据/span>N.据/span>是真正的根源据span class="katex"> P.据/span>(据/span>X据/span>)据/span>,然后韦达公式暗示据/P.>据P.>据span class="katex-display"> 一世据/span>=据/span>1据/span>σ.据/span>N.据/span>R.据/span>一世据/span>=据/span>-据/span>一种据/span>N.据/span>-据/span>1据/span>和据/span>1据/span>≤.据/span>一世据/span>据据/span>j据/span>≤.据/span>N.据/span>σ.据/span>R.据/span>一世据/span>R.据/span>j据/span>=据/span>一种据/span>N.据/span>-据/span>2据/span>那据/span>
这意味着据/P.>据P.>据span class="katex-display"> 一世据/span>=据/span>1据/span>σ.据/span>N.据/span>R.据/span>一世据/span>2据/span>=据/span>(据/span>一世据/span>=据/span>1据/span>σ.据/span>N.据/span>R.据/span>一世据/span>)据/span>2据/span>-据/span>2据/span>(据/span>1据/span>≤.据/span>一世据/span>据据/span>j据/span>≤.据/span>N.据/span>σ.据/span>R.据/span>一世据/span>R.据/span>j据/span>)据/span>=据/span>一种据/span>N.据/span>-据/span>1据/span>2据/span>-据/span>2据/span>一种据/span>N.据/span>-据/span>2据/span>≤.据/span>3.据/span>。据/span>
请注意,这也表明,据span class="katex"> 一种据/span>N.据/span>-据/span>1据/span>2据/span>-据/span>2据/span>一种据/span>N.据/span>-据/span>2据/span>≥据/span>0.据/span>因为左边是实数的平方和。自从据span class="katex"> 一种据/span>N.据/span>-据/span>1据/span>=据/span>±据/span>1据/span>, 这意味着据span class="katex"> 一种据/span>N.据/span>-据/span>2据/span>=据/span>-据/span>1据/span>。此外,通过韦达的公式,据span class="katex"> (据/span>R.据/span>1据/span>R.据/span>2据/span>⋯据/span>R.据/span>N.据/span>)据/span>2据/span>=据/span>1据/span>。然后由据一种target="_blank" rel="nofollow" href="//www.parkandroid.com/wiki/arithmetic-mean-geometric-mean/">AM-GM不等式据/一种>, 我们有据span class="katex"> σ.据/span>一世据/span>=据/span>1据/span>N.据/span>R.据/span>一世据/span>2据/span>≥据/span>N.据/span>,这意味着据span class="katex"> N.据/span>≤.据/span>3.据/span>。然后我们可以枚举所有此类多项式以找到解决方案据span class="katex"> X据/span>±据/span>1据/span>那据span class="katex"> X据/span>2据/span>±据/span>X据/span>-据/span>1据/span>, 和据span class="katex"> X据/span>3.据/span>-据/span>X据/span>±据/span>(据/span>X据/span>2据/span>-据/span>1据/span>)据/span>。据span class="katex"> □据/span>
让据span class="katex"> R.据/span>1据/span>那据/span>R.据/span>2据/span>和据span class="katex"> R.据/span>3.据/span>是多项式的根据span class="katex"> 5.据/span>X据/span>3.据/span>-据/span>1据/span>1据/span>X据/span>2据/span>+据/span>7.据/span>X据/span>+据/span>3.据/span>。评估据span class="katex"> R.据/span>1据/span>3.据/span>+据/span>R.据/span>2据/span>3.据/span>+据/span>R.据/span>3.据/span>3.据/span>。据/span>
在这个问题上,韦达定理的应用是不是很明显,而且表达必须改变。从据一种target="_blank" rel="nofollow" href="//www.parkandroid.com/assessment/techniques-trainer/advanced-factorization/">因式分解4据/一种>, 我们有据/P.>据P.>据span class="katex-display"> R.据/span>1据/span>3.据/span>+据/span>R.据/span>2据/span>3.据/span>+据/span>R.据/span>3.据/span>3.据/span>=据/span>3.据/span>R.据/span>1据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>[据/span>R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>-据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>]据/span>。据/span>
现在,我们只需要知道如何计算据span class="katex"> R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>。再次,从分解,我们有据/P.>据P.>据span class="katex-display"> R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>=据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>2据/span>-据/span>2据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>那据/span>
这使得我们可以得出这样的结论据/P.>据P.>据span class="katex-display"> R.据/span>1据/span>3.据/span>+据/span>R.据/span>2据/span>3.据/span>+据/span>R.据/span>3.据/span>3.据/span>=据/span>3.据/span>R.据/span>1据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>[据/span>R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>-据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>]据/span>=据/span>3.据/span>R.据/span>1据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>[据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>2据/span>-据/span>3.据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>]据/span>=据/span>3.据/span>×据/span>(据/span>-据/span>5.据/span>3.据/span>)据/span>+据/span>(据/span>5.据/span>1据/span>1据/span>)据/span>[据/span>(据/span>5.据/span>1据/span>1据/span>)据/span>2据/span>-据/span>3.据/span>×据/span>5.据/span>7.据/span>]据/span>=据/span>-据/span>1据/span>2据/span>5.据/span>4.据/span>9.据/span>。据/span>□据/span>
韦达的公式解决问题 - 基本据/h2>
让据span class="katex"> 一种据/span>和据span class="katex"> B.据/span>是2张不同的实数,且据span class="katex"> 一种据/span>2据/span>+据/span>3.据/span>一种据/span>+据/span>1据/span>=据/span>B.据/span>2据/span>+据/span>3.据/span>B.据/span>+据/span>1据/span>=据/span>0.据/span>。找到价值据span class="katex"> B.据/span>一种据/span>+据/span>一种据/span>B.据/span>。据/P.>据/div>
查找满足以下方程式系统的所有三倍的复数:据/P.>据P.>据span class="katex-display"> 一种据/span>+据/span>B.据/span>+据/span>C据/span>一种据/span>B.据/span>+据/span>B.据/span>C据/span>+据/span>C据/span>一种据/span>一种据/span>B.据/span>C据/span>=据/span>0.据/span>=据/span>0.据/span>=据/span>0.据/span>。据/span>
任何据span class="katex"> 3.据/span>数字据span class="katex"> (据/span>一种据/span>那据/span>B.据/span>那据/span>C据/span>)据/span>可以被认为是所述首一三次多项式的根据/P.>据P.>据span class="katex-display"> P.据/span>(据/span>X据/span>)据/span>=据/span>(据/span>X据/span>-据/span>一种据/span>)据/span>(据/span>X据/span>-据/span>B.据/span>)据/span>(据/span>X据/span>-据/span>C据/span>)据/span>。据/span>
应用韦达的公式多项式,我们得到据span class="katex"> P.据/span>(据/span>X据/span>)据/span>=据/span>X据/span>3.据/span>。因此,唯一的根据span class="katex"> P.据/span>(据/span>X据/span>)据/span>是据span class="katex"> X据/span>=据/span>0.据/span>(与多重据span class="katex"> 3.据/span>),这意味着据span class="katex"> (据/span>0.据/span>那据/span>0.据/span>那据/span>0.据/span>)据/span>是唯一的三重复数的一种满足式的系统。据span class="katex"> □据/span>
让据span class="katex"> R.据/span>1据/span>那据/span>R.据/span>2据/span>和据span class="katex"> R.据/span>3.据/span>是多项式的根据span class="katex"> 5.据/span>X据/span>3.据/span>-据/span>1据/span>1据/span>X据/span>2据/span>+据/span>7.据/span>X据/span>+据/span>3.据/span>。评估据/P.>据P.>据span class="katex-display"> R.据/span>1据/span>(据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>+据/span>R.据/span>2据/span>(据/span>1据/span>+据/span>R.据/span>3.据/span>+据/span>R.据/span>1据/span>)据/span>+据/span>R.据/span>3.据/span>(据/span>1据/span>+据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>)据/span>。据/span>
表达式等于据span class="katex"> R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>+据/span>2据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>。由韦达的公式,我们知道,据/P.>据P.>据span class="katex-display"> R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>=据/span>-据/span>5.据/span>-据/span>1据/span>1据/span>=据/span>5.据/span>1据/span>1据/span>那据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>=据/span>5.据/span>7.据/span>那据/span>R.据/span>1据/span>R.据/span>2据/span>R.据/span>3.据/span>=据/span>-据/span>5.据/span>3.据/span>那据/span>
所以据/P.>据P.>据span class="katex-display"> R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>+据/span>2据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>=据/span>5.据/span>1据/span>1据/span>+据/span>2据/span>×据/span>5.据/span>7.据/span>=据/span>5.据/span>2据/span>5.据/span>=据/span>5.据/span>。据/span>□据/span>
笔记据/strong>:一种常见的方法是尝试找到多项式的每个根源,特别是因为我们知道一个根源必须是真实的(为什么?)。但是,这不一定是一个可行的选择,因为我们很难确定实际的根源。据/em>
vieta的配方问题解决 - 中间据/h2>
让据span class="katex"> R.据/span>1据/span>那据/span>R.据/span>2据/span>和据span class="katex"> R.据/span>3.据/span>是多项式的根据span class="katex"> 5.据/span>X据/span>3.据/span>-据/span>1据/span>1据/span>X据/span>2据/span>+据/span>7.据/span>X据/span>+据/span>3.据/span>。评估据span class="katex"> R.据/span>1据/span>3.据/span>+据/span>R.据/span>2据/span>3.据/span>+据/span>R.据/span>3.据/span>3.据/span>。据/span>
在这个问题上,韦达定理的应用是不是很明显,而且表达必须改变。从分解,我们有据/P.>据P.>据span class="katex-display"> R.据/span>1据/span>3.据/span>+据/span>R.据/span>2据/span>3.据/span>+据/span>R.据/span>3.据/span>3.据/span>=据/span>3.据/span>R.据/span>1据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>[据/span>R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>-据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>]据/span>。据/span>
现在,我们只需要知道如何计算据span class="katex"> R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>。再次,从分解,我们有据/P.>据P.>据span class="katex-display"> R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>=据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>2据/span>-据/span>2据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>那据/span>
这使得我们可以得出这样的结论据/P.>据P.>据span class="katex-display"> R.据/span>1据/span>3.据/span>+据/span>R.据/span>2据/span>3.据/span>+据/span>R.据/span>3.据/span>3.据/span>=据/span>3.据/span>R.据/span>1据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>[据/span>R.据/span>1据/span>2据/span>+据/span>R.据/span>2据/span>2据/span>+据/span>R.据/span>3.据/span>2据/span>-据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>]据/span>=据/span>3.据/span>R.据/span>1据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>[据/span>(据/span>R.据/span>1据/span>+据/span>R.据/span>2据/span>+据/span>R.据/span>3.据/span>)据/span>2据/span>-据/span>3.据/span>(据/span>R.据/span>1据/span>R.据/span>2据/span>+据/span>R.据/span>2据/span>R.据/span>3.据/span>+据/span>R.据/span>3.据/span>R.据/span>1据/span>)据/span>]据/span>=据/span>3.据/span>×据/span>(据/span>-据/span>5.据/span>3.据/span>)据/span>+据/span>(据/span>5.据/span>1据/span>1据/span>)据/span>[据/span>(据/span>5.据/span>1据/span>1据/span>)据/span>2据/span>-据/span>3.据/span>×据/span>5.据/span>7.据/span>]据/span>=据/span>-据/span>1据/span>2据/span>5.据/span>4.据/span>9.据/span>。据/span>□据/span>
如果据span class="katex"> α.据/span>那据/span>β据/span>是根茎方程据span class="katex"> X据/span>2据/span>+据/span>2据/span>X据/span>+据/span>3.据/span>=据/span>0.据/span>,什么是二次方程式,其根据span class="katex"> (据/span>α.据/span>-据/span>α.据/span>1据/span>)据/span>2据/span>和据span class="katex"> (据/span>β据/span>-据/span>β据/span>1据/span>)据/span>2据/span>还是据/span>
[arml 2012,团队问题,#6]据/P.>据P.>的零据span class="katex"> F据/span>(据/span>X据/span>)据/span>=据/span>X据/span>6.据/span>+据/span>2据/span>X据/span>5.据/span>+据/span>3.据/span>X据/span>4.据/span>+据/span>5.据/span>X据/span>3.据/span>+据/span>8.据/span>X据/span>2据/span>+据/span>1据/span>3.据/span>X据/span>+据/span>2据/span>1据/span>是不同的复数。计算的平均值据span class="katex"> 一种据/span>+据/span>B.据/span>C据/span>+据/span>D.据/span>E.据/span>F据/span>超过所有可能的排列据span class="katex"> (据/span>一种据/span>那据/span>B.据/span>那据/span>C据/span>那据/span>D.据/span>那据/span>E.据/span>那据/span>F据/span>)据/span>这些六个号码。据/P.>据hr>
韦达的公式解决问题 - 高级据/h2>
[IMO名单]据B.R.>确定参数的所有实值据span class="katex"> 一种据/span>为此公式据/P.>据P.>据span class="katex-display"> 1据/span>6.据/span>X据/span>4.据/span>-据/span>一种据/span>X据/span>3.据/span>+据/span>(据/span>2据/span>一种据/span>+据/span>1据/span>7.据/span>)据/span>X据/span>2据/span>-据/span>一种据/span>X据/span>+据/span>1据/span>6.据/span>=据/span>0.据/span>
具有形成一几何级数恰好四个不同的实根。据/P.>据hr>
假设据span class="katex"> 一种据/span>满足了这个问题,这要求据span class="katex"> X据/span>那据span class="katex"> 问:据/span>X据/span>那据span class="katex"> 问:据/span>2据/span>X据/span>那据span class="katex"> 问:据/span>3.据/span>X据/span>是给定方程的根。然后据span class="katex"> X据/span>据/span>=据/span>0.据/span>我们可以假设据span class="katex"> |据/span>问:据/span>|据/span>>据/span>1据/span>, 以便据span class="katex"> |据/span>X据/span>|据/span>据据/span>|据/span>问:据/span>X据/span>|据/span>据据/span>|据/span>问:据/span>2据/span>X据/span>|据/span>据据/span>|据/span>问:据/span>3.据/span>X据/span>|据/span>。请注意,该系数是对称的,即,第一系数是相同的第五个,第二个是相同的第四,三是一样的三分之一。它保证了我们,如果据span class="katex"> α.据/span>是根,则其倒数(这是据span class="katex"> α.据/span>-据/span>1据/span>=据/span>α.据/span>1据/span>)也将是一个根目录。因此,据span class="katex"> X据/span>1据/span>=据/span>问:据/span>3.据/span>X据/span>那据/span>所以据span class="katex"> 问:据/span>=据/span>X据/span>-据/span>3.据/span>2据/span>和根据span class="katex"> X据/span>那据/span>X据/span>3.据/span>1据/span>那据/span>X据/span>3.据/span>-据/span>1据/span>那据/span>X据/span>-据/span>1据/span>。据/P.>据P.>现在,由韦达的公式,我们有据span class="katex"> X据/span>+据/span>X据/span>3.据/span>1据/span>+据/span>X据/span>3.据/span>-据/span>1据/span>+据/span>X据/span>-据/span>1据/span>=据/span>1据/span>6.据/span>一种据/span>和据span class="katex"> X据/span>3.据/span>4.据/span>+据/span>X据/span>3.据/span>2据/span>+据/span>1据/span>+据/span>1据/span>+据/span>X据/span>3.据/span>-据/span>2据/span>+据/span>X据/span>3.据/span>-据/span>4.据/span>=据/span>1据/span>6.据/span>2据/span>一种据/span>+据/span>1据/span>7.据/span>。在环境据span class="katex"> Z.据/span>=据/span>X据/span>3.据/span>1据/span>+据/span>X据/span>3.据/span>-据/span>1据/span>这些方程成为据/P.>据P.>据span class="katex-display"> Z.据/span>3.据/span>-据/span>2据/span>Z.据/span>=据/span>1据/span>6.据/span>一种据/span>那据/span>(据/span>Z.据/span>2据/span>-据/span>2据/span>)据/span>2据/span>+据/span>Z.据/span>2据/span>-据/span>2据/span>=据/span>1据/span>6.据/span>2据/span>一种据/span>+据/span>1据/span>7.据/span>。据/span>
替代据span class="katex"> 一种据/span>=据/span>1据/span>6.据/span>(据/span>Z.据/span>3.据/span>-据/span>2据/span>Z.据/span>)据/span>在第二个公式导致据span class="katex"> Z.据/span>4.据/span>-据/span>2据/span>Z.据/span>3.据/span>-据/span>3.据/span>Z.据/span>2据/span>+据/span>4.据/span>Z.据/span>+据/span>1据/span>6.据/span>1据/span>5.据/span>=据/span>0.据/span>。我们观察到,该多项式因素据span class="katex"> (据/span>Z.据/span>+据/span>2据/span>3.据/span>)据/span>(据/span>Z.据/span>-据/span>2据/span>5.据/span>)据/span>(据/span>Z.据/span>2据/span>-据/span>Z.据/span>-据/span>4.据/span>1据/span>)据/span>。自从据span class="katex"> |据/span>Z.据/span>|据/span>=据/span>|据/span>|据/span>|据/span>X据/span>3.据/span>1据/span>+据/span>X据/span>-据/span>3.据/span>1据/span>|据/span>|据/span>|据/span>≥据/span>2据/span>,唯一可行的价值是据span class="katex"> Z.据/span>=据/span>2据/span>5.据/span>。最后据span class="katex"> 一种据/span>=据/span>1据/span>7.据/span>0.据/span>。据/span>
整理上式,我们得到据span class="katex"> 1据/span>6.据/span>(据/span>X据/span>2据/span>+据/span>X据/span>2据/span>1据/span>)据/span>-据/span>1据/span>7.据/span>0.据/span>(据/span>X据/span>+据/span>X据/span>1据/span>)据/span>+据/span>3.据/span>5.据/span>7.据/span>=据/span>0.据/span>。为了简化计算,我们可以调用据span class="katex"> y据/span>=据/span>X据/span>+据/span>X据/span>1据/span>因此据span class="katex"> y据/span>2据/span>=据/span>X据/span>2据/span>+据/span>2据/span>+据/span>X据/span>2据/span>1据/span>,从而获得一种新形式据/P.>据P.>据span class="katex-display"> 1据/span>6.据/span>(据/span>y据/span>2据/span>-据/span>2据/span>)据/span>-据/span>1据/span>7.据/span>0.据/span>y据/span>+据/span>3.据/span>5.据/span>7.据/span>=据/span>0.据/span>⟹据/span>1据/span>6.据/span>y据/span>2据/span>-据/span>1据/span>7.据/span>0.据/span>y据/span>+据/span>3.据/span>2据/span>5.据/span>=据/span>0.据/span>那据/span>
其根源是据span class="katex"> 2据/span>5.据/span>和据span class="katex"> 8.据/span>6.据/span>5.据/span>。我们必须插上两回据span class="katex"> y据/span>=据/span>X据/span>+据/span>X据/span>1据/span>,导致我们两个二次方程式,最后得到据span class="katex-display"> 8.据/span>1据/span>那据/span>2据/span>1据/span>那据/span>2据/span>那据/span>8.据/span>。据/span>□据/span>
(由...制作据一种target="_blank" rel="nofollow" href="//www.parkandroid.com/profile/mark-4vl4ha/feed/">标记Hennings.据/一种>)据B.R.>显示据/P.>据P.>据span class="katex-display"> j据/span>=据/span>1据/span>σ.据/span>N.据/span>婴儿床据/span>2据/span>(据/span>2据/span>N.据/span>+据/span>1据/span>j据/span>π据/span>)据/span>=据/span>3.据/span>1据/span>N.据/span>(据/span>2据/span>N.据/span>-据/span>1据/span>)据/span>
对于任何整数据span class="katex"> N.据/span>≥据/span>1据/span>。据/P.>据hr>
(解决方案据一种target="_blank" rel="nofollow" href="//www.parkandroid.com/profile/aditya-85fq8c/">阿迪亚夏尔马据/一种>)据/P.>据P.>让我们先从德棣美弗定理任意正整数据span class="katex"> m据/span>:据/span>
(据/span>COS.据/span>X据/span>+据/span>一世据/span>罪据/span>X据/span>)据/span>m据/span>罪据/span>m据/span>X据/span>COS.据/span>m据/span>X据/span>+据/span>一世据/span>罪据/span>m据/span>X据/span>=据/span>COS.据/span>m据/span>X据/span>+据/span>一世据/span>罪据/span>m据/span>X据/span>=据/span>(据/span>婴儿床据/span>X据/span>+据/span>一世据/span>)据/span>m据/span>=据/span>婴儿床据/span>m据/span>X据/span>+据/span>(据/span>1据/span>m据/span>)据/span>婴儿床据/span>m据/span>-据/span>1据/span>(据/span>X据/span>)据/span>一世据/span>+据/span>⋯据/span>+据/span>(据/span>m据/span>-据/span>1据/span>m据/span>)据/span>婴儿床据/span>(据/span>X据/span>)据/span>一世据/span>m据/span>-据/span>1据/span>+据/span>一世据/span>m据/span>。据/span>
我们可以组RHS如下,因为我们有据span class="katex"> 一世据/span>2据/span>=据/span>-据/span>1据/span>:据/span>
RHS.据/span>=据/span>(据/span>婴儿床据/span>m据/span>X据/span>-据/span>(据/span>2据/span>m据/span>)据/span>婴儿床据/span>m据/span>-据/span>2据/span>X据/span>+据/span>⋯据/span>)据/span>+据/span>一世据/span>(据/span>(据/span>1据/span>m据/span>)据/span>婴儿床据/span>m据/span>-据/span>1据/span>X据/span>-据/span>(据/span>3.据/span>m据/span>)据/span>婴儿床据/span>m据/span>-据/span>3.据/span>X据/span>+据/span>⋯据/span>)据/span>。据/span>
等同于LHS和RHS虚部,我们得到据/P.>据P.>据span class="katex-display"> (据/span>1据/span>m据/span>)据/span>婴儿床据/span>m据/span>-据/span>1据/span>X据/span>-据/span>(据/span>3.据/span>m据/span>)据/span>婴儿床据/span>m据/span>-据/span>3.据/span>X据/span>+据/span>⋯据/span>=据/span>罪据/span>m据/span>X据/span>罪据/span>m据/span>X据/span>。据/span>
现在,如果我们让据span class="katex"> m据/span>=据/span>2据/span>N.据/span>+据/span>1据/span>和据span class="katex"> X据/span>=据/span>m据/span>j据/span>π据/span>对于任何一个据span class="katex"> j据/span>=据/span>1据/span>那据/span>2据/span>那据/span>......据/span>那据/span>m据/span>,则方程简化为据/P.>据P.>据span class="katex-display"> (据/span>1据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>婴儿床据/span>2据/span>N.据/span>X据/span>-据/span>(据/span>3.据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>婴儿床据/span>2据/span>N.据/span>-据/span>2据/span>X据/span>+据/span>⋯据/span>=据/span>0.据/span>
自从据span class="katex"> 罪据/span>m据/span>X据/span>=据/span>罪据/span>j据/span>π据/span>=据/span>0.据/span>。据/span>
让据span class="katex"> 婴儿床据/span>2据/span>X据/span>=据/span>你据/span>那据/span>然后等式在据span class="katex"> 你据/span>其根源之和由下式给出据span class="katex"> (据/span>1据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>(据/span>3.据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>因为我们知道从韦达的公式。自从据span class="katex"> X据/span>=据/span>2据/span>N.据/span>+据/span>1据/span>j据/span>π据/span>那据/span>
j据/span>=据/span>1据/span>σ.据/span>N.据/span>婴儿床据/span>2据/span>2据/span>N.据/span>+据/span>1据/span>j据/span>π据/span>=据/span>(据/span>1据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>(据/span>3.据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>=据/span>6.据/span>1据/span>2据/span>N.据/span>(据/span>2据/span>N.据/span>-据/span>1据/span>)据/span>=据/span>3.据/span>1据/span>N.据/span>(据/span>2据/span>N.据/span>-据/span>1据/span>)据/span>。据/span>□据/span>
什么是据/P.>据P.>据span class="katex-display"> K.据/span>=据/span>1据/span>σ.据/span>N.据/span>婴儿床据/span>4.据/span>(据/span>2据/span>N.据/span>+据/span>1据/span>K.据/span>π据/span>)据/span>还是据/span>
让据span class="katex"> W.据/span>是任何据一种href="//www.parkandroid.com/wiki/primitive-roots-of-unity/" class="wiki_link" title="原始" target="_blank">原始据/一种>据span class="katex"> (据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>TH.据/span>团结的根源。然后,据span class="katex"> P.据/span>(据/span>X据/span>)据/span>=据/span>X据/span>-据/span>1据/span>X据/span>2据/span>N.据/span>+据/span>1据/span>-据/span>1据/span>将有根源据span class="katex"> W.据/span>那据/span>W.据/span>2据/span>那据/span>......据/span>那据/span>W.据/span>2据/span>N.据/span>。我们也知道据span class="katex"> 罪据/span>(据/span>2据/span>N.据/span>+据/span>1据/span>π据/span>K.据/span>)据/span>=据/span>2据/span>一世据/span>W.据/span>K.据/span>/据/span>2据/span>-据/span>W.据/span>-据/span>K.据/span>/据/span>2据/span>和据span class="katex"> COS.据/span>(据/span>2据/span>N.据/span>+据/span>1据/span>π据/span>K.据/span>)据/span>=据/span>2据/span>W.据/span>K.据/span>/据/span>2据/span>+据/span>W.据/span>-据/span>K.据/span>/据/span>2据/span>。然后据span class="katex"> 婴儿床据/span>(据/span>2据/span>N.据/span>+据/span>1据/span>π据/span>K.据/span>)据/span>=据/span>W.据/span>K.据/span>-据/span>1据/span>一世据/span>(据/span>W.据/span>K.据/span>+据/span>1据/span>)据/span>。所以总和据span class="katex"> S.据/span>(说)是据/P.>据P.>据span class="katex-display"> S.据/span>=据/span>K.据/span>=据/span>1据/span>σ.据/span>N.据/span>(据/span>W.据/span>K.据/span>-据/span>1据/span>W.据/span>K.据/span>+据/span>1据/span>)据/span>4.据/span>。据/span>
现在,我们将通过做来做出根源的多项式转变据span class="katex"> y据/span>=据/span>X据/span>-据/span>1据/span>X据/span>+据/span>1据/span>, 然后据span class="katex"> X据/span>=据/span>y据/span>-据/span>1据/span>y据/span>+据/span>1据/span>。替换,在据span class="katex"> P.据/span>(据/span>X据/span>)据/span>:据/P.>据P.>据span class="katex-display"> P.据/span>(据/span>X据/span>)据/span>=据/span>y据/span>-据/span>1据/span>y据/span>+据/span>1据/span>-据/span>1据/span>(据/span>y据/span>-据/span>1据/span>y据/span>+据/span>1据/span>)据/span>2据/span>N.据/span>+据/span>1据/span>-据/span>1据/span>。据/span>
简化并使其成为一个多项式,我们得到据/P.>据P.>据span class="katex-display"> P.据/span>(据/span>y据/span>)据/span>=据/span>(据/span>y据/span>+据/span>1据/span>)据/span>2据/span>N.据/span>+据/span>1据/span>-据/span>(据/span>y据/span>-据/span>1据/span>)据/span>2据/span>N.据/span>+据/span>1据/span>。据/span>
扩展和再简化一下:据/P.>据P.>据span class="katex-display"> P.据/span>(据/span>y据/span>)据/span>=据/span>2据/span>(据/span>1据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>(据/span>y据/span>2据/span>)据/span>N.据/span>+据/span>2据/span>(据/span>3.据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>(据/span>y据/span>2据/span>)据/span>N.据/span>-据/span>1据/span>+据/span>2据/span>(据/span>5.据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>(据/span>y据/span>2据/span>)据/span>N.据/span>-据/span>2据/span>+据/span>⋯据/span>。据/span>
最后,由韦达定理,我们的总和据/P.>据P.>据span class="katex-display"> S.据/span>=据/span>K.据/span>=据/span>1据/span>σ.据/span>N.据/span>y据/span>K.据/span>4.据/span>=据/span>(据/span>K.据/span>=据/span>1据/span>σ.据/span>N.据/span>y据/span>K.据/span>2据/span>)据/span>2据/span>-据/span>2据/span>⎝据/span>⎛据/span>1据/span>≤.据/span>j据/span>据据/span>K.据/span>≤.据/span>N.据/span>σ.据/span>y据/span>j据/span>2据/span>y据/span>K.据/span>2据/span>⎠据/span>⎞据/span>=据/span>(据/span>2据/span>(据/span>1据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>-据/span>2据/span>(据/span>3.据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>)据/span>2据/span>-据/span>2据/span>(据/span>2据/span>(据/span>1据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>2据/span>(据/span>5.据/span>2据/span>N.据/span>+据/span>1据/span>)据/span>)据/span>=据/span>9.据/span>N.据/span>2据/span>(据/span>2据/span>N.据/span>-据/span>1据/span>)据/span>2据/span>-据/span>1据/span>5.据/span>N.据/span>(据/span>2据/span>N.据/span>-据/span>1据/span>)据/span>(据/span>N.据/span>-据/span>1据/span>)据/span>(据/span>2据/span>N.据/span>-据/span>3.据/span>)据/span>=据/span>4.据/span>5.据/span>N.据/span>(据/span>2据/span>N.据/span>-据/span>1据/span>)据/span>(据/span>4.据/span>N.据/span>2据/span>+据/span>1据/span>0.据/span>N.据/span>-据/span>9.据/span>)据/span>。据/span>□据/span>
让据span class="katex"> X据/span>1据/span>那据/span>X据/span>2据/span>那据/span>。据/span>。据/span>。据/span>那据/span>X据/span>1据/span>0.据/span>是多项式的根据span class="katex"> X据/span>1据/span>0.据/span>+据/span>X据/span>9.据/span>+据/span>⋯据/span>+据/span>X据/span>+据/span>1据/span>。找到价值据span class="katex-display"> N.据/span>=据/span>1据/span>σ.据/span>1据/span>0.据/span>1据/span>-据/span>X据/span>N.据/span>1据/span>
韦达根跳据/h2>
主要文章:据一种href="//www.parkandroid.com/wiki/vieta-root-jumping/" class="wiki_link" title="韦达根跳" target="_blank">韦达根跳据/一种>据/P.>据!-- end-meta -->
韦达跳跃是一种特殊的下降法,已成为在更高层次的数学奥赛数论问题颇为流行的昵称。像血统的其他情况下,当你必须解决(方程,同余或不等式或系统),一个丢番图方程时,其解决方案有一定的据一种href="//www.parkandroid.com/wiki/subroutines/" class="wiki_link" title="递归" target="_blank">递归据/一种>结构体。据/P.>据/div>
也可以看看据/h2>
硕士概念如此据/h4>
问题加载......据/P.>据P.Class="note-text">注意加载......据/P.>据P.Class="set-text">设置加载......据/P.>据/div>