(9) $\sqrt{8} - \frac{1}{\sqrt{2}}$ (10) $\frac{9}{\sqrt{3}} - \sqrt{6} \times \sqrt{2}$ (11) $\sqrt{3} (\sqrt{6} + \sqrt{3}) - \frac{8}{\sqrt{2}}$ (12) $(\sqrt{2} + 2)^2 - \sqrt{72}$

わかりました。画像にある4つの数学の問題を解きます。

1. 問題の内容

(9) $\sqrt{8} - \frac{1}{\sqrt{2}}$

(10) $\frac{9}{\sqrt{3}} - \sqrt{6} \times \sqrt{2}$

(11) $\sqrt{3} (\sqrt{6} + \sqrt{3}) - \frac{8}{\sqrt{2}}$

(12) $(\sqrt{2} + 2)^2 - \sqrt{72}$

2. 解き方の手順

(9)

$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$

$\frac{1}{\sqrt{2}} = \frac{1 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{2}}{2}$

$2\sqrt{2} - \frac{\sqrt{2}}{2} = \frac{4\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2}$

(10)

$\frac{9}{\sqrt{3}} = \frac{9\sqrt{3}}{\sqrt{3}\sqrt{3}} = \frac{9\sqrt{3}}{3} = 3\sqrt{3}$

$\sqrt{6} \times \sqrt{2} = \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

$3\sqrt{3} - 2\sqrt{3} = \sqrt{3}$

(11)

$\sqrt{3} (\sqrt{6} + \sqrt{3}) = \sqrt{3}\sqrt{6} + \sqrt{3}\sqrt{3} = \sqrt{18} + 3 = \sqrt{9 \times 2} + 3 = 3\sqrt{2} + 3$

$\frac{8}{\sqrt{2}} = \frac{8\sqrt{2}}{\sqrt{2}\sqrt{2}} = \frac{8\sqrt{2}}{2} = 4\sqrt{2}$

$3\sqrt{2} + 3 - 4\sqrt{2} = 3 - \sqrt{2}$

(12)

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

$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$

$6 + 4\sqrt{2} - 6\sqrt{2} = 6 - 2\sqrt{2}$

3. 最終的な答え

(9) $\frac{3\sqrt{2}}{2}$

(10) $\sqrt{3}$

(11) $3 - \sqrt{2}$

(12) $6 - 2\sqrt{2}$