与えられた式を簡略化します。 $\frac{x-y}{x+y} + \frac{2xy}{x^2-y^2}$代数学式の簡略化分数式因数分解代数2025/4/131. 問題の内容与えられた式を簡略化します。x−yx+y+2xyx2−y2\frac{x-y}{x+y} + \frac{2xy}{x^2-y^2}x+yx−y+x2−y22xy2. 解き方の手順まず、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)次に、共通分母 (x+y)(x−y)(x+y)(x-y)(x+y)(x−y) を使って、二つの分数を足し合わせます。x−yx+y+2xyx2−y2=x−yx+y+2xy(x+y)(x−y)\frac{x-y}{x+y} + \frac{2xy}{x^2-y^2} = \frac{x-y}{x+y} + \frac{2xy}{(x+y)(x-y)}x+yx−y+x2−y22xy=x+yx−y+(x+y)(x−y)2xyx−yx+y⋅x−yx−y+2xy(x+y)(x−y)=(x−y)2(x+y)(x−y)+2xy(x+y)(x−y)\frac{x-y}{x+y} \cdot \frac{x-y}{x-y} + \frac{2xy}{(x+y)(x-y)} = \frac{(x-y)^2}{(x+y)(x-y)} + \frac{2xy}{(x+y)(x-y)}x+yx−y⋅x−yx−y+(x+y)(x−y)2xy=(x+y)(x−y)(x−y)2+(x+y)(x−y)2xy(x−y)2+2xy(x+y)(x−y)=x2−2xy+y2+2xy(x+y)(x−y)=x2+y2(x+y)(x−y)\frac{(x-y)^2 + 2xy}{(x+y)(x-y)} = \frac{x^2 - 2xy + y^2 + 2xy}{(x+y)(x-y)} = \frac{x^2 + y^2}{(x+y)(x-y)}(x+y)(x−y)(x−y)2+2xy=(x+y)(x−y)x2−2xy+y2+2xy=(x+y)(x−y)x2+y2x2+y2x2−y2\frac{x^2 + y^2}{x^2 - y^2}x2−y2x2+y23. 最終的な答えx2+y2x2−y2\frac{x^2 + y^2}{x^2 - y^2}x2−y2x2+y2