$x = \sqrt{7} + \sqrt{2}$、 $y = \sqrt{7} - \sqrt{2}$ のとき、$x^2 - y^2$ の値を求めよ。代数学式の計算平方根因数分解2025/8/61. 問題の内容x=7+2x = \sqrt{7} + \sqrt{2}x=7+2、 y=7−2y = \sqrt{7} - \sqrt{2}y=7−2 のとき、x2−y2x^2 - y^2x2−y2 の値を求めよ。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 をそれぞれ計算します。x+y=(7+2)+(7−2)=7+7+2−2=27x+y = (\sqrt{7} + \sqrt{2}) + (\sqrt{7} - \sqrt{2}) = \sqrt{7} + \sqrt{7} + \sqrt{2} - \sqrt{2} = 2\sqrt{7}x+y=(7+2)+(7−2)=7+7+2−2=27x−y=(7+2)−(7−2)=7+2−7+2=22x-y = (\sqrt{7} + \sqrt{2}) - (\sqrt{7} - \sqrt{2}) = \sqrt{7} + \sqrt{2} - \sqrt{7} + \sqrt{2} = 2\sqrt{2}x−y=(7+2)−(7−2)=7+2−7+2=22求めた x+yx+yx+y と x−yx-yx−y を (x+y)(x−y)(x+y)(x-y)(x+y)(x−y) に代入します。(x+y)(x−y)=(27)(22)=47×2=414(x+y)(x-y) = (2\sqrt{7})(2\sqrt{2}) = 4\sqrt{7 \times 2} = 4\sqrt{14}(x+y)(x−y)=(27)(22)=47×2=4143. 最終的な答え4144\sqrt{14}414