$x + \frac{1}{x} = \sqrt{2} + 1$ のとき、$x^2 + \frac{1}{x^2}$ の値を求める問題です。代数学式の計算二次式有理化2025/6/31. 問題の内容x+1x=2+1x + \frac{1}{x} = \sqrt{2} + 1x+x1=2+1 のとき、x2+1x2x^2 + \frac{1}{x^2}x2+x21 の値を求める問題です。2. 解き方の手順まず、x+1x=2+1x + \frac{1}{x} = \sqrt{2} + 1x+x1=2+1 の両辺を2乗します。(x+1x)2=(2+1)2(x + \frac{1}{x})^2 = (\sqrt{2} + 1)^2(x+x1)2=(2+1)2x2+2(x)(1x)+1x2=2+22+1x^2 + 2(x)(\frac{1}{x}) + \frac{1}{x^2} = 2 + 2\sqrt{2} + 1x2+2(x)(x1)+x21=2+22+1x2+2+1x2=3+22x^2 + 2 + \frac{1}{x^2} = 3 + 2\sqrt{2}x2+2+x21=3+22x2+1x2=3+22−2x^2 + \frac{1}{x^2} = 3 + 2\sqrt{2} - 2x2+x21=3+22−2x2+1x2=1+22x^2 + \frac{1}{x^2} = 1 + 2\sqrt{2}x2+x21=1+223. 最終的な答え1+221 + 2\sqrt{2}1+22