绝对值不平等据/h1>
已经有一个帐户?据一种href="//www.parkandroid.com/account/login/?next=/wiki/absolute-value-inequalities/" class="ax-click" data-ax-id="clicked_signup_modal_login" data-ax-type="link">这里登录。据/a>
相关......据/h4>
- 代数据/span>>据/span>
- 代数据/span>>据/span>
绝对值不等式据/strong>处理不平等据span class="katex">
(据/span>据据/span>那据/span>≤.据/span>那据/span>>据/span>那据/span>≥据/span>)据/span>与表达据一种href="//www.parkandroid.com/wiki/absolute-value/" class="wiki_link" title="绝对值GydF4y2Ba" target="_blank">绝对值据/a>标志。例如,据/p>
求解不等式据span class="katex">
X据/span>:据span class="katex">
|据/span>X据/span>-据/span>9.据/span>|据/span>据据/span>4.据/span>。据/span>
我们将使用定义据一种href="//www.parkandroid.com/wiki/absolute-value/" class="wiki_link" title="绝对值GydF4y2Ba" target="_blank">绝对值据/a>解决这个问题,这是据p>
-据/span>4.据/span>据据/span>X据/span>-据/span>9.据/span>5.据/span>据据/span>X据/span>据据/span>4.据/span>据据/span>1据/span>3.据/span>。据/span>□据/span>
解决这样的问题时,最简单的准则是考虑分段函数,并把它分割成相关情况:据/p>
|据/span>X据/span>-据/span>一种据/span>|据/span>=据/span>{据/span>-据/span>X据/span>+据/span>一种据/span>X据/span>-据/span>一种据/span>为了据/span>X据/span>据据/span>一种据/span>为了据/span>X据/span>≥据/span>一种据/span>。据/span>
然而,存在多种技术来解决这种不平等喜欢使用基本属性,考虑的情况下,图形可视化等,这wiki页面解释了所有详细的工作实例和问题,试图沿着这些技术。这些技术从解决基本的不平等进一步延伸到更难绝对值不等式开始。据/p>
简介和基本属性据/h2>
回想一下据一种href="//www.parkandroid.com/wiki/absolute-value/" class="wiki_link" title="绝对值GydF4y2Ba" target="_blank">绝对值据/a>:据/p>
对于任何实数据span class="katex"> X据/span>, 它的据strong>绝对值据/strong>被定义为据/p>
|据/span>X据/span>|据/span>=据/span>⎩据/span>⎪据/span>⎨据/span>⎪据/span>⎧据/span>X据/span>0.据/span>-据/span>X据/span>如果据/span>X据/span>>据/span>0.据/span>如果据/span>X据/span>=据/span>0.据/span>如果据/span>X据/span>据据/span>0.据/span>。据/span>
功能据span class="katex"> F.据/span>(据/span>X据/span>)据/span>=据/span>|据/span>X据/span>|据/span>也被称为模函数。据span class="katex"> □据/span>
让我们据span class="katex">
X据/span>是一个变量或一个代数表达式,并让据span class="katex">
一种据/span>是一个真实的数字据span class="katex">
一种据/span>>据/span>0.据/span>。然后以下不等式持有:据/p>
对于据span class="katex">
F.据/span>(据/span>X据/span>)据/span>和据span class="katex">
G据/span>(据/span>X据/span>)据/span>那据/span>函数据span class="katex">
X据/span>那据/span>我们有据/p>
1据/span>。据/span>|据/span>F.据/span>(据/span>X据/span>)据/span>|据/span>≤.据/span>G据/span>(据/span>X据/span>)据/span>⇔据/span>-据/span>G据/span>(据/span>X据/span>)据/span>≤.据/span>F.据/span>(据/span>X据/span>)据/span>≤.据/span>G据/span>(据/span>X据/span>)据/span>2据/span>。据/span>|据/span>F.据/span>(据/span>X据/span>)据/span>|据/span>≥据/span>G据/span>(据/span>X据/span>)据/span>⇔据/span>F.据/span>(据/span>X据/span>)据/span>≤.据/span>-据/span>G据/span>(据/span>X据/span>)据/span>或者据/span>F.据/span>(据/span>X据/span>)据/span>≥据/span>G据/span>(据/span>X据/span>)据/span>3.据/span>。据/span>|据/span>F.据/span>(据/span>X据/span>)据/span>|据/span>据据/span>G据/span>(据/span>X据/span>)据/span>⇔据/span>-据/span>G据/span>(据/span>X据/span>)据/span>据据/span>F.据/span>(据/span>X据/span>)据/span>据据/span>G据/span>(据/span>X据/span>)据/span>4.据/span>。据/span>|据/span>F.据/span>(据/span>X据/span>)据/span>|据/span>>据/span>G据/span>(据/span>X据/span>)据/span>⇔据/span>F.据/span>(据/span>X据/span>)据/span>据据/span>-据/span>G据/span>(据/span>X据/span>)据/span>或者据/span>F.据/span>(据/span>X据/span>)据/span>>据/span>G据/span>(据/span>X据/span>)据/span>。据/span> 以下是对不等式绝对值的其他一些属性:据/p>
归纳上述两个不等式,我们得到据/p>
|据/span>±据/span>一种据/span>1据/span>±据/span>一种据/span>2据/span>±据/span>一种据/span>3.据/span>±据/span>⋯据/span>±据/span>一种据/span>N据/span>|据/span>≤.据/span>|据/span>一种据/span>1据/span>|据/span>+据/span>|据/span>一种据/span>2据/span>|据/span>+据/span>⋯据/span>+据/span>|据/span>一种据/span>N据/span>|据/span>。据/span> 让我们来看一个例子,看看如何应用这个公式。据/p>
找到的最小值据span class="katex">
|据/span>X据/span>-据/span>1据/span>|据/span>+据/span>|据/span>X据/span>-据/span>2据/span>|据/span>+据/span>⋯据/span>+据/span>|据/span>X据/span>-据/span>1据/span>0.据/span>0.据/span>|据/span>。据/p>
使用上面的不平等,我们尝试删除变量据span class="katex">
X据/span>经过据/p>
|据/span>X据/span>-据/span>1据/span>|据/span>+据/span>|据/span>X据/span>-据/span>2据/span>|据/span>+据/span>⋯据/span>+据/span>|据/span>X据/span>-据/span>1据/span>0.据/span>0.据/span>|据/span>≥据/span>|据/span>(据/span>X据/span>-据/span>1据/span>)据/span>+据/span>(据/span>X据/span>-据/span>2据/span>)据/span>+据/span>⋯据/span>+据/span>(据/span>X据/span>-据/span>5.据/span>0.据/span>)据/span>-据/span>(据/span>X据/span>-据/span>5.据/span>1据/span>)据/span>-据/span>(据/span>X据/span>-据/span>5.据/span>2据/span>)据/span>-据/span>⋯据/span>-据/span>(据/span>X据/span>-据/span>1据/span>0.据/span>0.据/span>)据/span>|据/span>=据/span>2据/span>5.据/span>0.据/span>0.据/span>。据/span> 为了验证这是一个有效的情况下,我们必须有据span class="katex">
{据/span>(据/span>X据/span>-据/span>1据/span>)据/span>那据/span>(据/span>X据/span>-据/span>2据/span>)据/span>那据/span>......据/span>那据/span>(据/span>X据/span>-据/span>5.据/span>0.据/span>)据/span>}据/span>≥据/span>0.据/span>和据span class="katex">
{据/span>(据/span>X据/span>-据/span>5.据/span>1据/span>)据/span>那据/span>(据/span>X据/span>-据/span>5.据/span>2据/span>)据/span>那据/span>......据/span>那据/span>(据/span>X据/span>-据/span>1据/span>0.据/span>0.据/span>)据/span>}据/span>≤.据/span>0.据/span>。这减少了只有两种情况:据/p>
{据/span>X据/span>-据/span>5.据/span>0.据/span>X据/span>-据/span>5.据/span>1据/span>≥据/span>0.据/span>≤.据/span>0.据/span>。据/span> 因此,获得的最小值时据span class="katex">
X据/span>∈据/span>[据/span>5.据/span>0.据/span>那据/span>5.据/span>1据/span>]据/span>。据span class="katex">
□据/span>
这是一个尝试的问题。据/p>
如果解决方案设置不平等的制度据span class="katex-display">
{据/span>|据/span>一种据/span>+据/span>1据/span>|据/span>据据/span>3.据/span>|据/span>B.据/span>-据/span>1据/span>|据/span>据据/span>1据/span>0.据/span>是据span class="katex">
X据/span>据据/span>一种据/span>+据/span>B.据/span>据据/span>y据/span>那据/span>那是什么据span class="katex">
X据/span>和据span class="katex">
y据/span>还是据/span>
解决问题的技术 - 案例工作据/h2>
在这里,我们看看取代数字的绝对值的定义。为了“撤消”绝对值标志,我们可以获得正值或负值,因为绝对值据span class="katex">
-据/span>5.据/span>与绝对值相同据span class="katex">
5.据/span>,这是据span class="katex">
5.据/span>。这将成为我们在那里有多种情况的方法。据/p>
的主要步骤(用于处理线性/多个直线型绝对值不等式)是据/p>
我们将探讨如何在未来的3个例子做到这一点。据/p>
解决据span class="katex">
|据/span>X据/span>+据/span>3.据/span>|据/span>据据/span>7.据/span>。据/p>
情况1:据/strong>
X据/span>+据/span>3.据/span>是非负面的,还是据span class="katex">
X据/span>⩾据/span>-据/span>3.据/span> 案例2:据/strong>
X据/span>+据/span>3.据/span>是负的,或据span class="katex">
X据/span>据据/span>-据/span>3.据/span> 最后,我们采取了这些不等式的联盟,因为他们彼此独立:据span class="katex-display">
-据/span>1据/span>0.据/span>据据/span>X据/span>据据/span>4.据/span>。据/span>□据/span>
解决据span class="katex">
2据/span>|据/span>X据/span>+据/span>2据/span>|据/span>-据/span>|据/span>X据/span>+据/span>5.据/span>|据/span>⩽据/span>4.据/span>。据/p>
情况1:据/strong>
X据/span>+据/span>2据/span>>据/span>0.据/span>和据/span>X据/span>+据/span>5.据/span>>据/span>0.据/span>⟹据/span>X据/span>>据/span>-据/span>2据/span> 案例2:据/strong>
X据/span>+据/span>2据/span>>据/span>0.据/span>和据/span>X据/span>+据/span>5.据/span>⩽据/span>0.据/span> 案例3:据/strong>
X据/span>+据/span>2据/span>⩽据/span>0.据/span>和据/span>X据/span>+据/span>5.据/span>>据/span>0.据/span>⟹据/span>-据/span>5.据/span>据据/span>X据/span>⩽据/span>-据/span>2据/span> 案例4:据/strong>
X据/span>+据/span>2据/span>⩽据/span>和据/span>X据/span>+据/span>5.据/span>⩽据/span>⟹据/span>X据/span>⩽据/span>-据/span>5.据/span> 因此,从案例1和3,我们有据span class="katex-display">
3.据/span>-据/span>1据/span>3.据/span>⩽据/span>X据/span>⩽据/span>5.据/span>。据/span>□据/span>
解决据span class="katex">
|据/span>X据/span>2据/span>+据/span>4.据/span>X据/span>+据/span>4.据/span>|据/span>-据/span>2据/span>|据/span>X据/span>+据/span>2据/span>|据/span>+据/span>1据/span>据据/span>2据/span>1据/span>。据/p>
我们得到了很多很多这里的案件。还是我们?据/p>
重写给出的不等式据span class="katex-display">
|据/span>|据/span>(据/span>X据/span>+据/span>2据/span>)据/span>2据/span>|据/span>|据/span>-据/span>2据/span>|据/span>X据/span>+据/span>2据/span>|据/span>+据/span>1据/span>据据/span>2据/span>1据/span>。据/span>不应该据span class="katex">
(据/span>X据/span>+据/span>2据/span>)据/span>2据/span>永远保持乐观?是的。我们不需要在那里签名。据/p>
我们只剩下两种情况 - 一个在哪里据span class="katex">
X据/span>+据/span>2据/span>(在绝对值标志中)是正的,另一个是否定的。据/p>
情况1:据/strong>
X据/span>+据/span>2据/span>≥据/span>0.据/span>⟹据/span>X据/span>≥据/span>-据/span>2据/span> 我们知道,不平等将等于零时,整个表达式的一个因素等于零,即在据span class="katex">
-据/span>1据/span>-据/span>2据/span>
1据/span>和据span class="katex">
-据/span>1据/span>+据/span>2据/span>
1据/span>,所以我们想知道每个因素是否在特定值中是正的或负数据span class="katex">
X据/span>是小于据span class="katex">
-据/span>1据/span>-据/span>2据/span>
1据/span>, 比...更棒据span class="katex">
-据/span>1据/span>-据/span>2据/span>
1据/span>和据/strong>少于据span class="katex">
-据/span>1据/span>+据/span>2据/span>
1据/span>和大于-1 +据span class="katex">
2据/span>
1据/span>。我们还使用属性负负次给予正面的,积极的次正给正,负的时间正给负。为了满足该左边小于零,我们寻找据span class="katex">
X据/span>这给我们左侧的负值。据/p>
我们可以看到,二次不等式的左边小于零的时候据span class="katex">
-据/span>1据/span>-据/span>2据/span>
1据/span>据据/span>X据/span>据据/span>-据/span>1据/span>+据/span>2据/span>
1据/span>。据/span>(据/span>1据/span>)据/span> 案例2:据/strong>
X据/span>+据/span>2据/span>据据/span>0.据/span>⟹据/span>X据/span>据据/span>-据/span>2据/span>
绝对值的财产告诉我们据span class="katex">
|据/span>一种据/span>|据/span>=据/span>一种据/span>对于非负据span class="katex">
一种据/span>,所以在这种情况下,据span class="katex-display">
|据/span>X据/span>+据/span>3.据/span>|据/span>据据/span>7.据/span>⟹据/span>X据/span>+据/span>3.据/span>据据/span>7.据/span>⟹据/span>X据/span>据据/span>4.据/span>。据/span>现在,我们这里有两个不等式,并且这种情况下的解决方案既不平等的交集。这是因为据span class="katex">
X据/span>应该满足既作为它们依赖彼此的据span class="katex">
(据/span>只因为据span class="katex">
X据/span>≥据/span>-据/span>3.据/span>我们有据span class="katex">
X据/span>据据/span>4.据/span>)据/span>。因此,对于这种情况下的解决方案是据span class="katex">
-据/span>3.据/span>≤.据/span>X据/span>据据/span>4.据/span>。据/p>
再次,据span class="katex">
|据/span>一种据/span>|据/span>=据/span>-据/span>一种据/span>为负值据span class="katex">
一种据/span>, 所以据span class="katex-display">
|据/span>X据/span>+据/span>3.据/span>|据/span>据据/span>7.据/span>⟹据/span>-据/span>(据/span>X据/span>+据/span>3.据/span>)据/span>据据/span>7.据/span>⟹据/span>X据/span>>据/span>-据/span>1据/span>0.据/span>。据/span>对于与上述相同的原因,我们应该采取两个不等式的交点,这是据span class="katex">
-据/span>1据/span>0.据/span>据据/span>X据/span>据据/span>-据/span>3.据/span>。据/p>
在这种情况下,我们有据span class="katex-display">
2据/span>(据/span>X据/span>+据/span>2据/span>)据/span>-据/span>(据/span>X据/span>+据/span>5.据/span>)据/span>2据/span>X据/span>+据/span>4.据/span>-据/span>X据/span>-据/span>5.据/span>X据/span>-据/span>1据/span>X据/span>⇒据/span>-据/span>2据/span>据据/span>X据/span>⩽据/span>4.据/span>⩽据/span>4.据/span>⩽据/span>4.据/span>⩽据/span>5.据/span>⩽据/span>5.据/span>。据/span>
因为它总是如此据span class="katex">
X据/span>+据/span>5.据/span>>据/span>X据/span>+据/span>2据/span>那据/span>这种情况下是不可能的。据/p>
在这种情况下,我们有据span class="katex-display">
-据/span>2据/span>(据/span>X据/span>+据/span>2据/span>)据/span>-据/span>(据/span>X据/span>+据/span>5.据/span>)据/span>-据/span>2据/span>X据/span>-据/span>4.据/span>-据/span>X据/span>-据/span>5.据/span>-据/span>3.据/span>X据/span>-据/span>9.据/span>-据/span>3.据/span>X据/span>X据/span>⇒据/span>3.据/span>-据/span>1据/span>3.据/span>⩽据/span>X据/span>⩽据/span>4.据/span>⩽据/span>4.据/span>⩽据/span>4.据/span>⩽据/span>1据/span>3.据/span>⩾据/span>3.据/span>-据/span>1据/span>3.据/span>⩽据/span>-据/span>2据/span>。据/span>
在这种情况下,我们有据span class="katex-display">
-据/span>2据/span>(据/span>X据/span>+据/span>2据/span>)据/span>+据/span>(据/span>X据/span>+据/span>5.据/span>)据/span>-据/span>2据/span>X据/span>-据/span>4.据/span>+据/span>X据/span>+据/span>5.据/span>-据/span>X据/span>+据/span>1据/span>-据/span>X据/span>X据/span>⩽据/span>4.据/span>⩽据/span>4.据/span>⩽据/span>4.据/span>⩽据/span>3.据/span>⩾据/span>3.据/span>那据/span>这是不可能的,因为据span class="katex">
X据/span>⩽据/span>-据/span>5.据/span>。据/span>
在这种情况下,我们有据span class="katex-display">
(据/span>X据/span>+据/span>2据/span>)据/span>2据/span>-据/span>2据/span>(据/span>X据/span>+据/span>2据/span>)据/span>+据/span>1据/span>(据/span>X据/span>+据/span>2据/span>-据/span>1据/span>)据/span>2据/span>(据/span>X据/span>+据/span>1据/span>)据/span>2据/span>(据/span>X据/span>+据/span>1据/span>)据/span>2据/span>-据/span>2据/span>1据/span>(据/span>X据/span>+据/span>1据/span>+据/span>2据/span>
1据/span>)据/span>(据/span>X据/span>+据/span>1据/span>-据/span>2据/span>
1据/span>)据/span>据据/span>2据/span>1据/span>据据/span>2据/span>1据/span>据据/span>2据/span>1据/span>据据/span>0.据/span>据据/span>0.据/span>。据/span>为了解决这个二次不等式,我们可以用一个标志分析图。据/p>
(据/span>请注意,这些不平等满足据span class="katex">
X据/span>≥据/span>-据/span>2据/span>。据/span>)据/span>
在这种情况下,我们有据span class="katex-display">
(据/span>X据/span>+据/span>2据/span>)据/span>2据/span>+据/span>2据/span>(据/span>X据/span>+据/span>2据/span>)据/span>+据/span>1据/span>(据/span>X据/span>+据/span>2据/span>+据/span>1据/span>)据/span>2据/span>(据/span>X据/span>+据/span>3.据/span>)据/span>2据/span>(据/span>X据/span>+据/span>3.据/span>)据/span>2据/span>-据/span>2据/span>1据/span>(据/span>X据/span>+据/span>3.据/span>+据/span>2据/span>
1据/span>)据/span>(据/span>X据/span>+据/span>3.据/span>-据/span>2据/span>