# 分母の有理化（分母が和または差の式のとき）

## 分母の有理化の基本

$\dfrac{c}{\sqrt{a}}=\dfrac{c\sqrt{a}}{\sqrt{a}\sqrt{a}}=\dfrac{c\sqrt{a}}{a}$

1次の式の分母を有理化しなさい。

(1)　$\dfrac{5}{\sqrt{2}}$

(2)　$\dfrac{3}{\sqrt{7}}$

(3)　$\dfrac{1}{\sqrt{3}}$

(4)　$\dfrac{15}{2\sqrt{5}}$

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1

(1)
$\dfrac{5}{\sqrt{2}}=\dfrac{5\sqrt{2}}{\sqrt{2}\sqrt{2}}=\dfrac{5\sqrt{2}}{2}$

(2)
$\dfrac{3}{\sqrt{7}}=\dfrac{3\sqrt{7}}{\sqrt{7}\sqrt{7}}=\dfrac{3\sqrt{7}}{7}$

(3)
$\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{\sqrt{3}\sqrt{3}}=\dfrac{\sqrt{3}}{3}$

(4)
$\dfrac{15}{2\sqrt{5}}=\dfrac{15\sqrt{5}}{2\sqrt{5}\sqrt{5}}=\dfrac{15\sqrt{5}}{10}=\dfrac{3\sqrt{5}}{2}$

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## 分母の有理化の応用（分母が和または差の式）

$\dfrac{c}{\sqrt{a}+\sqrt{b}}=\dfrac{c(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}=\dfrac{c(\sqrt{a}-\sqrt{b})}{a-b}$

$\dfrac{c}{\sqrt{a}-\sqrt{b}}=\dfrac{c(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}=\dfrac{c(\sqrt{a}+\sqrt{b})}{a-b}$

2次の式の分母を有理化しなさい。

(1)　$\dfrac{3}{\sqrt{5}-\sqrt{2}}$

(2)　$\dfrac{2}{\sqrt{7}+\sqrt{3}}$

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(1)
$\dfrac{3}{\sqrt{5}-\sqrt{2}}=\dfrac{3(\sqrt{5}+\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})}=\dfrac{3(\sqrt{5}+\sqrt{2})}{3}=\sqrt{5}+\sqrt{2}$

(2)
$\dfrac{2}{\sqrt{7}+\sqrt{3}}=\dfrac{2(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}=\dfrac{2(\sqrt{7}-\sqrt{3})}{4}=\dfrac{\sqrt{7}-\sqrt{3}}{2}$

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