画像にある数式問題のうち、以下の3問を解きます。 (16) $(\sqrt{5} + 7\sqrt{2})(\sqrt{5} - \sqrt{2})$ (18) $(\sqrt{6} + 2)^2 - (\sqrt{6} - 2)^2$ (19) $\sqrt{2}(2\sqrt{2}-4) - \frac{4}{\sqrt{2}(\sqrt{2}-2)}$

1. 問題の内容

画像にある数式問題のうち、以下の3問を解きます。

(16) $(\sqrt{5} + 7\sqrt{2})(\sqrt{5} - \sqrt{2})$

(18) $(\sqrt{6} + 2)^2 - (\sqrt{6} - 2)^2$

(19) $\sqrt{2}(2\sqrt{2}-4) - \frac{4}{\sqrt{2}(\sqrt{2}-2)}$

2. 解き方の手順

(16) $(\sqrt{5} + 7\sqrt{2})(\sqrt{5} - \sqrt{2})$ を展開します。

$(\sqrt{5} + 7\sqrt{2})(\sqrt{5} - \sqrt{2}) = (\sqrt{5})^2 - \sqrt{5}\sqrt{2} + 7\sqrt{2}\sqrt{5} - 7(\sqrt{2})^2$

$= 5 - \sqrt{10} + 7\sqrt{10} - 7 \times 2 = 5 + 6\sqrt{10} - 14 = 6\sqrt{10} - 9$

(18) $(\sqrt{6} + 2)^2 - (\sqrt{6} - 2)^2$ を展開します。

$(\sqrt{6} + 2)^2 = (\sqrt{6})^2 + 2 \times \sqrt{6} \times 2 + 2^2 = 6 + 4\sqrt{6} + 4 = 10 + 4\sqrt{6}$

$(\sqrt{6} - 2)^2 = (\sqrt{6})^2 - 2 \times \sqrt{6} \times 2 + 2^2 = 6 - 4\sqrt{6} + 4 = 10 - 4\sqrt{6}$

$(\sqrt{6} + 2)^2 - (\sqrt{6} - 2)^2 = (10 + 4\sqrt{6}) - (10 - 4\sqrt{6}) = 10 + 4\sqrt{6} - 10 + 4\sqrt{6} = 8\sqrt{6}$

(19) $\sqrt{2}(2\sqrt{2}-4) - \frac{4}{\sqrt{2}(\sqrt{2}-2)}$ を計算します。

$\sqrt{2}(2\sqrt{2}-4) = 2\sqrt{2}\sqrt{2} - 4\sqrt{2} = 2 \times 2 - 4\sqrt{2} = 4 - 4\sqrt{2}$

$\frac{4}{\sqrt{2}(\sqrt{2}-2)} = \frac{4}{\sqrt{2}\sqrt{2} - 2\sqrt{2}} = \frac{4}{2 - 2\sqrt{2}} = \frac{4}{2 - 2\sqrt{2}} \times \frac{2 + 2\sqrt{2}}{2 + 2\sqrt{2}} = \frac{4(2 + 2\sqrt{2})}{2^2 - (2\sqrt{2})^2} = \frac{8 + 8\sqrt{2}}{4 - 8} = \frac{8 + 8\sqrt{2}}{-4} = -2 - 2\sqrt{2}$

$\sqrt{2}(2\sqrt{2}-4) - \frac{4}{\sqrt{2}(\sqrt{2}-2)} = (4 - 4\sqrt{2}) - (-2 - 2\sqrt{2}) = 4 - 4\sqrt{2} + 2 + 2\sqrt{2} = 6 - 2\sqrt{2}$

3. 最終的な答え

(16) $6\sqrt{10} - 9$

(18) $8\sqrt{6}$

(19) $6 - 2\sqrt{2}$