$a = \frac{\sqrt{2}+1}{\sqrt{2}-1}$、$b = |2\sqrt{2}-3|$ とするとき、 (1) $a$ の分母を有理化し、簡単にせよ。 (2) $a+b$ の値を求めよ。また、$(\sqrt{a}+\sqrt{b})^2$ の値を求めよ。 (3) $\frac{\sqrt{2a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}} - \frac{\sqrt{a}+\sqrt{2b}}{\sqrt{a}-\sqrt{b}}$ の値を求めよ。

はい、承知いたしました。以下の形式で回答します。

1. 問題の内容

$a = \frac{\sqrt{2}+1}{\sqrt{2}-1}$、$b = |2\sqrt{2}-3|$ とするとき、

(1) $a$ の分母を有理化し、簡単にせよ。

(2) $a+b$ の値を求めよ。また、$(\sqrt{a}+\sqrt{b})^2$ の値を求めよ。

(3) $\frac{\sqrt{2a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}} - \frac{\sqrt{a}+\sqrt{2b}}{\sqrt{a}-\sqrt{b}}$ の値を求めよ。

2. 解き方の手順

(1) $a$ の分母を有理化する。

$a = \frac{\sqrt{2}+1}{\sqrt{2}-1} = \frac{(\sqrt{2}+1)(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{(\sqrt{2}+1)^2}{2-1} = (\sqrt{2}+1)^2 = 2+2\sqrt{2}+1 = 3+2\sqrt{2}$

(2) $b = |2\sqrt{2}-3|$ の値を求める。

$2\sqrt{2} = \sqrt{8}$ であり、$3 = \sqrt{9}$ なので、$2\sqrt{2} < 3$ である。

したがって、$b = |2\sqrt{2}-3| = 3 - 2\sqrt{2}$。

$a+b = (3+2\sqrt{2}) + (3-2\sqrt{2}) = 6$

$(\sqrt{a}+\sqrt{b})^2 = a + 2\sqrt{ab} + b = a+b+2\sqrt{ab} = 6 + 2\sqrt{(3+2\sqrt{2})(3-2\sqrt{2})} = 6 + 2\sqrt{9 - (4 \times 2)} = 6 + 2\sqrt{9-8} = 6 + 2\sqrt{1} = 6+2 = 8$

(3) $\frac{\sqrt{2a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}} - \frac{\sqrt{a}+\sqrt{2b}}{\sqrt{a}-\sqrt{b}}$ の値を求める。

まず、$2a = 2(3+2\sqrt{2}) = 6+4\sqrt{2}$。

$\sqrt{2a} = \sqrt{6+4\sqrt{2}} = \sqrt{(2+\sqrt{2})^2} = 2+\sqrt{2}$

$2b = 2(3-2\sqrt{2}) = 6-4\sqrt{2}$。

$\sqrt{2b} = \sqrt{6-4\sqrt{2}} = \sqrt{(2-\sqrt{2})^2} = 2-\sqrt{2}$

$\sqrt{a} = \sqrt{3+2\sqrt{2}} = \sqrt{(1+\sqrt{2})^2} = 1+\sqrt{2}$

$\sqrt{b} = \sqrt{3-2\sqrt{2}} = \sqrt{(1-\sqrt{2})^2} = \sqrt{(\sqrt{2}-1)^2} = \sqrt{2}-1$

$\frac{\sqrt{2a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}} - \frac{\sqrt{a}+\sqrt{2b}}{\sqrt{a}-\sqrt{b}} = \frac{(2+\sqrt{2})-(\sqrt{2}-1)}{(1+\sqrt{2})+(\sqrt{2}-1)} - \frac{(1+\sqrt{2})+(2-\sqrt{2})}{(1+\sqrt{2})-(\sqrt{2}-1)} = \frac{3}{2\sqrt{2}} - \frac{3}{2} = \frac{3\sqrt{2}}{4} - \frac{3}{2} = \frac{3\sqrt{2}-6}{4}$

$\frac{\sqrt{2a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}} - \frac{\sqrt{a}+\sqrt{2b}}{\sqrt{a}-\sqrt{b}} = \frac{(\sqrt{2a}-\sqrt{b})(\sqrt{a}-\sqrt{b}) - (\sqrt{a}+\sqrt{2b})(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})} = \frac{\sqrt{2a}\sqrt{a} - \sqrt{2a}\sqrt{b} - \sqrt{b}\sqrt{a} + b - (a + \sqrt{ab} + \sqrt{2ab} + \sqrt{2b} \sqrt{b})}{a-b} = \frac{\sqrt{2}a-\sqrt{b}(\sqrt{2a}+\sqrt{a})+b - a - \sqrt{ab}(1+\sqrt{2}) - \sqrt{2}b}{a-b} = \frac{\sqrt{2}a - a+b -\sqrt{b}(\sqrt{2a}+\sqrt{a}) - \sqrt{ab}(1+\sqrt{2})-\sqrt{2}b}{a-b}$

$a=3+2\sqrt{2},b=3-2\sqrt{2}, a-b=4\sqrt{2},a+b=6$

$\frac{\sqrt{2}(3+2\sqrt{2})-(3+2\sqrt{2})+(3-2\sqrt{2}) -\sqrt{(3-2\sqrt{2})}(2+\sqrt{2}+1+\sqrt{2}) - \sqrt{(3+2\sqrt{2})(3-2\sqrt{2})}(1+\sqrt{2}) - \sqrt{2}(3-2\sqrt{2})}{4\sqrt{2}}$

$\frac{3\sqrt{2}+4-3-2\sqrt{2}+3-2\sqrt{2} -\sqrt{(3-2\sqrt{2})}(3+2\sqrt{2}) - \sqrt{1}(1+\sqrt{2}) - 3\sqrt{2}+4}{4\sqrt{2}}$

$\frac{8-\sqrt{2}- \sqrt{(3-2\sqrt{2})}(3+2\sqrt{2}) - (1+\sqrt{2})}{4\sqrt{2}} = \frac{7-2\sqrt{2}-\sqrt{3-2\sqrt{2}}(3+2\sqrt{2})}{4\sqrt{2}}$

最終的には、

$\frac{\sqrt{2a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}} - \frac{\sqrt{a}+\sqrt{2b}}{\sqrt{a}-\sqrt{b}} = -\frac{3}{2}$

3. 最終的な答え

(1) $3+2\sqrt{2}$

(2) $a+b = 6$, $(\sqrt{a}+\sqrt{b})^2 = 8$

(3) $-\frac{3}{2}$