$x = \sqrt{2} + 1$、 $y = \sqrt{2} - 1$ のとき、$x^2 - y^2$ の値を求める問題です。代数学因数分解式の計算平方根2025/5/231. 問題の内容x=2+1x = \sqrt{2} + 1x=2+1、 y=2−1y = \sqrt{2} - 1y=2−1 のとき、x2−y2x^2 - y^2x2−y2 の値を求める問題です。2. 解き方の手順x2−y2x^2 - y^2x2−y2 を因数分解します。x2−y2=(x+y)(x−y)x^2 - y^2 = (x+y)(x-y)x2−y2=(x+y)(x−y)x+yx+yx+y と x−yx-yx−y を計算します。x+y=(2+1)+(2−1)=2+1+2−1=22x + y = (\sqrt{2} + 1) + (\sqrt{2} - 1) = \sqrt{2} + 1 + \sqrt{2} - 1 = 2\sqrt{2}x+y=(2+1)+(2−1)=2+1+2−1=22x−y=(2+1)−(2−1)=2+1−2+1=2x - y = (\sqrt{2} + 1) - (\sqrt{2} - 1) = \sqrt{2} + 1 - \sqrt{2} + 1 = 2x−y=(2+1)−(2−1)=2+1−2+1=2したがって、x2−y2=(x+y)(x−y)=(22)(2)=42x^2 - y^2 = (x+y)(x-y) = (2\sqrt{2})(2) = 4\sqrt{2}x2−y2=(x+y)(x−y)=(22)(2)=423. 最終的な答え424\sqrt{2}42