$\frac{\sqrt{2} + 1}{\sqrt{2} - 1}$ を計算する問題です。算数式の計算有理化平方根2025/5/151. 問題の内容2+12−1\frac{\sqrt{2} + 1}{\sqrt{2} - 1}2−12+1 を計算する問題です。2. 解き方の手順分母に無理数があるので、分母を有理化します。分母の 2−1\sqrt{2} - 12−1 の共役な無理数 2+1\sqrt{2} + 12+1 を分子と分母に掛けます。2+12−1=(2+1)(2+1)(2−1)(2+1)\frac{\sqrt{2} + 1}{\sqrt{2} - 1} = \frac{(\sqrt{2} + 1)(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)}2−12+1=(2−1)(2+1)(2+1)(2+1)分子を展開します。(2+1)(2+1)=(2)2+2(2)(1)+12=2+22+1=3+22(\sqrt{2} + 1)(\sqrt{2} + 1) = (\sqrt{2})^2 + 2(\sqrt{2})(1) + 1^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2}(2+1)(2+1)=(2)2+2(2)(1)+12=2+22+1=3+22分母を展開します。(2−1)(2+1)=(2)2−12=2−1=1(\sqrt{2} - 1)(\sqrt{2} + 1) = (\sqrt{2})^2 - 1^2 = 2 - 1 = 1(2−1)(2+1)=(2)2−12=2−1=1よって、(2+1)(2+1)(2−1)(2+1)=3+221=3+22\frac{(\sqrt{2} + 1)(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{3 + 2\sqrt{2}}{1} = 3 + 2\sqrt{2}(2−1)(2+1)(2+1)(2+1)=13+22=3+223. 最終的な答え3+223 + 2\sqrt{2}3+22