与えられた分数の引き算 $\frac{x+y}{x-y} - \frac{x-y}{x+y}$ を計算し、結果をできるだけ簡潔な形にすること。代数学分数式の計算通分展開因数分解2025/6/161. 問題の内容与えられた分数の引き算 x+yx−y−x−yx+y\frac{x+y}{x-y} - \frac{x-y}{x+y}x−yx+y−x+yx−y を計算し、結果をできるだけ簡潔な形にすること。2. 解き方の手順まず、二つの分数を共通の分母で通分します。共通の分母は (x−y)(x+y)(x-y)(x+y)(x−y)(x+y) です。x+yx−y−x−yx+y=(x+y)(x+y)(x−y)(x+y)−(x−y)(x−y)(x−y)(x+y)\frac{x+y}{x-y} - \frac{x-y}{x+y} = \frac{(x+y)(x+y)}{(x-y)(x+y)} - \frac{(x-y)(x-y)}{(x-y)(x+y)}x−yx+y−x+yx−y=(x−y)(x+y)(x+y)(x+y)−(x−y)(x+y)(x−y)(x−y)次に、分子を展開します。(x+y)(x+y)=x2+2xy+y2(x+y)(x+y) = x^2 + 2xy + y^2(x+y)(x+y)=x2+2xy+y2(x−y)(x−y)=x2−2xy+y2(x-y)(x-y) = x^2 - 2xy + y^2(x−y)(x−y)=x2−2xy+y2したがって、(x+y)(x+y)(x−y)(x+y)−(x−y)(x−y)(x−y)(x+y)=x2+2xy+y2−(x2−2xy+y2)(x−y)(x+y)\frac{(x+y)(x+y)}{(x-y)(x+y)} - \frac{(x-y)(x-y)}{(x-y)(x+y)} = \frac{x^2 + 2xy + y^2 - (x^2 - 2xy + y^2)}{(x-y)(x+y)}(x−y)(x+y)(x+y)(x+y)−(x−y)(x+y)(x−y)(x−y)=(x−y)(x+y)x2+2xy+y2−(x2−2xy+y2)分子の引き算を行います。x2+2xy+y2−(x2−2xy+y2)=x2+2xy+y2−x2+2xy−y2=4xyx^2 + 2xy + y^2 - (x^2 - 2xy + y^2) = x^2 + 2xy + y^2 - x^2 + 2xy - y^2 = 4xyx2+2xy+y2−(x2−2xy+y2)=x2+2xy+y2−x2+2xy−y2=4xy分母を展開します。(x−y)(x+y)=x2−y2(x-y)(x+y) = x^2 - y^2(x−y)(x+y)=x2−y2したがって、4xyx2−y2\frac{4xy}{x^2 - y^2}x2−y24xy3. 最終的な答え4xyx2−y2\frac{4xy}{x^2 - y^2}x2−y24xy