合理化分母

分母合理化就是把分数的底部移动到顶部。 $\压裂{1}{\ sqrt {2}}$例如，它的分母是无理数。我们很快就会看到它等于 $\压裂{\ sqrt {2}} {2}$．我们来看看分母无理数的分数。

考虑分数

$\压裂{一}{\ sqrt {b}}。$

完全可以写成

${对齐}\ \开始压裂{一}{\ sqrt {b}} & = \压裂{一}{\ sqrt {b}} \ * 1 \ \ & = \压裂{一}大概{b}}{\ \ * \压裂{\ sqrt {b}} {\ sqrt {b}} \ \ & = \压裂{\倍根号{b}} {b}。结束\{对齐}$

我们现在已经把分母合理化了，大多数问题都将依赖于这个简单的技术。

使分母合理化 $\压裂{1}{\ sqrt{2}}。$

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运用我们所学到的，我们拥有了

$\压裂{1}{\ sqrt{2}} = \压裂{1}{\ sqrt{2}} \ * \压裂{\ sqrt {2}} {\ sqrt{2}} = \压裂{\ sqrt{2}}{2}。\ _ \广场$

使分母合理化 $\frac{2}{\sqrt {\sqrt {2}}}$．

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我们有

${对齐}\ \开始dfrac {2} {\ sqrt {\ sqrt {2}}} & = \ dfrac{2×\√{\ sqrt {2}}} {\ sqrt {\ sqrt{2}}×\ sqrt {\ sqrt {2 }}}\\ & = \ dfrac{2 \√{\ sqrt{2}}}{\√6 {2 }}\\ & = \ dfrac{2 \√{\ sqrt{2}}×\ sqrt {2}} {\ sqrt{2}×\ sqrt {2 }}\\ & = \ dfrac{2 \√{2 \√{2}}}{2}\ \ & = \√6{2 \√{2}}。\ _\方形\端{对齐}$

使分母合理化 $\压裂{\ sqrt {y-1}} {\ sqrt {y + 1}}$．

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我们有

$\开始{对齐}\ dfrac {\ sqrt {y-1}} {\ sqrt {y + 1}} & = \ dfrac{\大大概{y-1} \大)(\ \大\√{y + 1} \大)}{\大\√{y + 1} \大)大\ \√{y + 1} \大)}\ \ & = \ dfrac {\ sqrt {y ^ 2 - 1}} {y + 1}。\ _\方形\端{对齐}$

使表达合理化 $\frac{y^2 \√{x^2 - 1}}{\√{x+1}}。$

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分母有理化，得到

$\压裂{y ^ 2 \√{x ^ 2 - 1}} {\ sqrt {x + 1}} \ * \压裂{\ sqrt {x + 1}} {\ sqrt {x + 1}} = \压裂{y ^ 2 \√x ^ 2 {1} \ * \ sqrt {x + 1}} {x + 1}。$

进一步简化，我们有

${对齐}\ \开始压裂{y ^ 2 \√{(x - 1) (x + 1)} \ * \ sqrt {x + 1}} {x + 1} & = \压裂{y ^ 2 \√{(x - 1) (x + 1) ^ 2}} {x + 1} \ \ & = \压裂{y ^ 2 x + 1 \√{x}} {x + 1}。结束\{对齐}$

取消了 $x + 1,$最终结果是

$y ^ 2 \√{x - 1}。\ _ \广场$

使分母合理化 $\压裂{3}{\ sqrt [3] {3}}$

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我们有

${对齐}\ \开始dfrac {3} {\ sqrt [3] {3}} & = \ dfrac{3×\ sqrt [3] {9}} {\ sqrt[3]{3}×\ sqrt [3] {9 }}\\ & = \ dfrac{3×\ sqrt [3] {9}} {\ sqrt [3] {27 }}\\ & = \ dfrac{3×\ sqrt [3] {9}} {3} \ \ & = \ sqrt[3]{9}。\ _\方形\端{对齐}$

现在考虑表单的分数 $\frac{a}{b + \sqrt{c}}，$在哪里 $一个$， $b$而且 $c$都是整数。如果我们乘以 $c \压裂{b + \ sqrt {}} {b + \ sqrt {c}},$我们将获得 $\frac{ab +a\√{c}}{b^2 + 2b\√{c} + c}，$你可以看到，这是一个更复杂的分数形式。因此，每当我们得到这种类型的分数，我们利用平方之差公式 $(x-y)(x+y)=x²-y²。$这个特别有用，因为如果 $X = \根号{m}$而且 $y = \ sqrt {n},$然后 $大\ \√大概{n} {m} - \ \大)大\ \√{m} + \ sqrt {n} \大)= m - n。$现在我们准备通过乘以来对原表达式的分母进行理性化 $\压裂{b - \ sqrt {c}} {b - \ sqrt {c}}:$ $\压裂{一}{b + \ sqrt {c}} \ * \压裂{b - \ sqrt {c}} {b - \ sqrt {c}} = \压裂{ab - \ sqrt {c}} {b ^ 2 - c}。$

简化下面的表达式:

$\压裂{3}{\ sqrt{3}} + \压裂{2}{\ sqrt {2} + \ sqrt{5}}。$

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让我们分别讨论这两个组成部分。 $\压裂{3}{\ sqrt {3}}$可以通过合理化简化 $\压裂{\ sqrt {3}} {\ sqrt {3}}$右边的部分 $\frac{2}{\√{2}+ \√{5}}$可以通过乘以 $\压裂大概{2}- {\ \ sqrt {5}} {\ sqrt {2} - \ sqrt {5}}:$ $\压裂{3}{\ sqrt{3}} \ * \压裂{\ sqrt {3}} {\ sqrt{3}} + \压裂{2}{\ sqrt大概{5}}{2}+ \ \ * \压裂大概{2}- {\ \ sqrt {5}} {\ sqrt {2} - \ sqrt{5}} = \压裂{3 \ sqrt{3}}{3} + \压裂{2 \√{2}- \ sqrt{5})}{3}。$由此我们得到了答案 $3 \ \压裂{sqrt{3} 2 \√{2}+ 2 \√{5}}{3}。\ _ \广场$

简化下面的表达式: $7 \ \压裂{sqrt {6} + \ sqrt{303}}{大概{3}}\ \ * \√大概{98}{101}- \)。$

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我们从写作开始 $\ sqrt {98}$作为 $7 \ sqrt {2}:$ $7 \ \压裂{sqrt {6} + \ sqrt{303}}{大概{3}}\ \ * \ sqrt{101} - 7 \√{2})。$分母理性化 $\ sqrt {3}$给了 $\begin{aligned} \frac{7\√{6}\times \√{3}+ \\√{303}\√{3}}{3}\times(\ \√{101}- 7\√{2})=& \frac{7\√{3\√2 \times 3}}{3} \times(\ \√{101}- 7\√{2})\\ =& \frac{7\√{3^2 \times 2} + \√{3^2 \times101}}{3} \times(\√{101}- 7\√{2})\\ =& frac{3\times 7\√{2}+ 3\√101}}{3}\times(\√{101}- 7\√{2})。结束\{对齐}$约掉 $3.$左边的给 $\{对齐}开始7 \√{2}+ \ sqrt {101}) \ * \ sqrt{101} - 7 \√{2})& = 7 \ sqrt {202} - 98 + 101 - 7 \ sqrt{202} \ \ & = 3。\ _\方形\端{对齐}$

简化

$大\ dfrac{\{\√{5}\大)}^ 3 -{\大(大概{4}\ \大)}^ 3}{{\大(大概{5}\ \大)}^ 3 +{\大(大概{4}\ \大)}^ 3}。$

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我们有

${对齐}\ \开始dfrac{{\√{5})}^ 3 - {(\ sqrt{4})} ^ 3}{{\√{5})}^ 3 +{\√{4})}^ 3}& = \ dfrac{5 - \ \√√6 4)5 + \√{20}+ 4)}{\√5 + \√4)5 - \√{20}+ 4 )}\\ & = \ dfrac{\ \√大大概{4}{5}- \ \大)、大(9 + 2 \√{5}\大)}{\大\√大概{4}{5}+ \ \大)大(9 - 2 \ \ sqrt{5} \大 )}\\ & = \ dfrac{\ \√大大概{4}{5}- \ \大)、大(9 + 2 \√{5}\大)\大(9 + 2 \√{5}\大)}{\大\√大概{4}{5}+ \ \大)\大9 - 2 \√{5}\大)大(9 + 2 \ \ sqrt{5} \大 )}\\ & = \ dfrac{\ \√大{5}-2大)\ \大(81 + 20 + 36 \ sqrt{5} \大)}{\大\√{5}+ 2 \大)(81 - 20 )}\\ & = \ dfrac{\大(大概5 - 2 \ \大)\大(101 + 36 \ sqrt{5} \大)}{\大(大概5 + 2 \ \大)(61 )}\\ & = \ 大(101 + 36 dfrac {\ \ sqrt{5} \大)、大(大概5 - 2 \ \大)}{(61)\ \√5 + 2大\大 )}\\ & = \ dfrac{\大(101 + 36 \ sqrt{5} \大)、大(大概5 - 2 \ \大)\大(大概5 - 2 \ \大)}{(61)大\ \√5 + 2 \大)\ \√5 - 2大\大 )}\\ & = \ 大(101 + 36 dfrac {\ \ sqrt{5} \大)、大(5 + 4 - 4 \√5 \大)}{(61)(5 - 4 )}\\ & = \ dfrac{\大(101 + 36 \ sqrt{5} \大)\大大概5(9 - 4 \ \大)}{61}\ \ & = \ dfrac{189 - 80 \倍根号5}{61}。\ _\方形\端{对齐}$

使分母合理化 $大概{\ dfrac {{\ \ sqrt [3] {2}}}} {{\ sqrt [4] {\ sqrt {5}}} - {\ sqrt [4] {\ sqrt{3}}}}。$

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我们有

${对齐}\ \开始dfrac{{\√6 {\ sqrt [3] {2}}}} {{\ sqrt [4] {\ sqrt {5}}} - {\ sqrt[4]{\√6 {3 }}}} & = \ dfrac{\离开(2 ^{\压裂{1}{3}}\右)^{\压裂{1}{2}}}{{\离开(5 ^{\压裂{1}{4}}\右)^{\压裂{1}{2}}- \离开(3 ^{\压裂{1}{4}}\右)^{\压裂{1}{2 }}}}\\ & = \ dfrac{2 ^{\压裂{1}{6}}}{5 ^{\压裂{1}{8}}- 3 ^{\压裂{1}{8 }}}\\ & = \ dfrac{2 ^{\压裂{1}{6}}\离开(5 ^{\压裂{1}{8}}+ 3 ^{\压裂{1}{8}}\右)}{\离开(5 ^{\压裂{1}{8}}- 3 ^{\压裂{1}{8}}\)\离开(5 ^{\压裂{1}{8}}+ 3 ^{\压裂{1}{8}}\右)}\ \ & =\ dfrac{2 ^{\压裂{1}{6}}\离开(5 ^{\压裂{1}{8}}+ 3 ^{\压裂{1}{8}}\)\离开(5 ^{\压裂{1}{4}}+ 3 ^{\压裂{1}{4}}\右)}{\离开(5 ^{\压裂{1}{4}}- 3 ^{\压裂{1}{4}}\)\离开(5 ^{\压裂{1}{4}}+ 3 ^{\压裂{1}{4}}\ )}\\ & = \ dfrac{2 ^{\压裂{1}{6}}\离开(5 ^{\压裂{1}{8}}+ 3 ^{\压裂{1}{8}}\)\离开(5 ^{\压裂{1}{4}}+ 3 ^{\压裂{1}{4}}\右)}{\离开(5 ^{\压裂{1}{2}}- 3 ^{\压裂{1}{2}}\ )}\\ & = \ dfrac{2 ^{\压裂{1}{6}}\离开(5 ^{\压裂{1}{8}}+ 3 ^{\压裂{1}{8}}\)\离开(5 ^{\压裂{1}{4}}+3 ^{\压裂{1}{4}}\)\离开(5 ^{\压裂{1}{2}}+ 3 ^{\压裂{1}{2}}\右)}{\离开(5 ^{\压裂{1}{2}}- 3 ^{\压裂{1}{2}}\)\离开(5 ^{\压裂{1}{2}}+ 3 ^{\压裂{1}{2}}\ )}\\ & = \ dfrac{2 ^{\压裂{1}{6}}\离开(5 ^{\压裂{1}{8}}+ 3 ^{\压裂{1}{8}}\)\离开(5 ^{\压裂{1}{4}}+ 3 ^{\压裂{1}{4}}\)\离开(5 ^{\压裂{1}{2}}+ 3 ^{\压裂{1}{2}}\右)}{5 - 3}\ \ & = \ dfrac{2 ^{\压裂{1}{6}}\离开(5 ^{\压裂{1}{8}}+ 3 ^{\压裂{1}{8}}\)\离开(5 ^{\压裂{1}{4}}+ 3 ^{\压裂{1}{4}}\)\离开(5 ^{\压裂{1}{2}}+3 ^{\压裂{1}{2}}\右)}{2}。\ _\方形\端{对齐}$

使分母合理化 $\ dfrac {1} {\ sqrt [5] {\ sqrt [4] {\ sqrt [3] {\ sqrt{2}}}}}。$

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我们有

${对齐}\ \开始dfrac {1} {\ sqrt [5] {\ sqrt大概[3][4]{\{\√6 {2 }}}}} & = \ 左\√dfrac{1×{\ [5]{\ sqrt [4] {\ sqrt [3] {\ sqrt{2}}}} \右)}^ {4}}{\ sqrt [5] {\ sqrt [4] {\ sqrt [3] {\ sqrt{2}}}}×{\左\√[5]{\ sqrt [4] {\ sqrt [3] {\ sqrt{2}}}} \右)}^ {4 }}\\ & = \ dfrac {\ sqrt [5] {\ sqrt [3] {\ sqrt {2}}}} {\ sqrt大概[3][4]{\{\√6 {2 }}}}\\ & = \ dfrac {\ sqrt [5] {\ sqrt [3] {\ sqrt{2}}}×{\左\√[4]{\ sqrt [3] {\ sqrt{2}}} \右)}^ {3}}{\ sqrt [4] {\ sqrt [3] {\ sqrt{2}}}×{\左\√[4]{\ sqrt [3] {\ sqrt{2}}} \右)}^ {3}}\ \ & =\ dfrac {\ sqrt [5] {\ sqrt [3] {\ sqrt{2}}}×\ sqrt [4] {\ sqrt {2}}} {\ sqrt [3] {\ sqrt {2 }}}\\ & = \ dfrac {\ sqrt [5] {\ sqrt [3] {\ sqrt{2}}}×\ sqrt [4] {\ sqrt{2}}×{\左\√[3]{\ sqrt{2}} \右)}^ {2}}{\ sqrt [3] {\ sqrt{2}}×{\左\√[3]{\ sqrt{2}} \右)}^ {2 }}\\ & = \ dfrac {\ sqrt [5] {\ sqrt [3] {\ sqrt{2}}}×\ sqrt [4] {\ sqrt{2}}×\ sqrt[3]{2}}{\√6 {2 }}\\ & = \ dfrac {\ sqrt [5] {\ sqrt [3] {\ sqrt{2}}}×\ sqrt [4] {\ sqrt{2}}×\ sqrt[3]{2}×\ sqrt {2}} {\ sqrt{2}×\ sqrt {2}} \ \ & = \ dfrac {\ sqrt [5] {\ sqrt [3] {\ sqrt{2}}}×\根号[4]{\根号{2}}× \根号[3]{2}× \根号{2}}{2}。\ _\方形\端{对齐}$

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