次の5つの式を計算する問題です。 (1) $(1+\sqrt{2}+\sqrt{3})^2$ (2) $(\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}+\sqrt{3}-\sqrt{5})$ (3) $\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{32}}$ (4) $\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{2+\sqrt{3}}{2-\sqrt{3}}$ (5) $\frac{1}{1+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + \frac{1}{\sqrt{3}+2}$

1. 問題の内容

次の5つの式を計算する問題です。

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

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

(3) $\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{32}}$

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

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

2. 解き方の手順

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

展開して計算します。

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

$= 1 + \sqrt{2} + \sqrt{3} + \sqrt{2} + 2 + \sqrt{6} + \sqrt{3} + \sqrt{6} + 3$

$= 1 + 2 + 3 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6}$

$= 6 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6}$

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

$A = \sqrt{2}+\sqrt{3}$とおくと、$(A+\sqrt{5})(A-\sqrt{5}) = A^2 - (\sqrt{5})^2 = A^2 - 5$

$(\sqrt{2}+\sqrt{3})^2 = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}$

よって、$(5 + 2\sqrt{6}) - 5 = 2\sqrt{6}$

(3) $\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{32}}$

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

$\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$

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

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

それぞれを有理化します。

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

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

よって、$(7 - 4\sqrt{3}) + (7 + 4\sqrt{3}) = 14$

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

それぞれを有理化します。

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

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

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

よって、$(-1+\sqrt{2}) + (-\sqrt{2}+\sqrt{3}) + (-\sqrt{3}+2) = -1+2 = 1$

3. 最終的な答え

(1) $6 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6}$

(2) $2\sqrt{6}$

(3) $\frac{3\sqrt{2}}{8}$

(4) $14$

(5) $1$