$\frac{\sqrt{2} + 1}{\sqrt{2} - 1}$ の値を求めます。算数有理化平方根計算2025/5/81. 問題の内容2+12−1\frac{\sqrt{2} + 1}{\sqrt{2} - 1}2−12+1 の値を求めます。2. 解き方の手順分母を有理化するために、分母と分子に 2+1\sqrt{2} + 12+1 を掛けます。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)分母は (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2(a−b)(a+b)=a2−b2 の公式を使って計算できます。(2−1)(2+1)=(2)2−(1)2=2−1=1(\sqrt{2} - 1)(\sqrt{2} + 1) = (\sqrt{2})^2 - (1)^2 = 2 - 1 = 1(2−1)(2+1)=(2)2−(1)2=2−1=1分子は (2+1)2=(2)2+2(2)(1)+12=2+22+1=3+22(\sqrt{2} + 1)^2 = (\sqrt{2})^2 + 2(\sqrt{2})(1) + 1^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2}(2+1)2=(2)2+2(2)(1)+12=2+22+1=3+22したがって、(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