与えられた分数の式を簡略化します。式は次の通りです。 $\frac{\frac{1}{x+y} - \frac{1}{x-y}}{\frac{1}{x+y} + \frac{1}{x-y}}$代数学分数式式の簡略化代数2025/6/31. 問題の内容与えられた分数の式を簡略化します。式は次の通りです。1x+y−1x−y1x+y+1x−y\frac{\frac{1}{x+y} - \frac{1}{x-y}}{\frac{1}{x+y} + \frac{1}{x-y}}x+y1+x−y1x+y1−x−y12. 解き方の手順まず、分子と分母をそれぞれ計算します。分子:1x+y−1x−y=(x−y)−(x+y)(x+y)(x−y)=x−y−x−y(x+y)(x−y)=−2y(x+y)(x−y)=−2yx2−y2\frac{1}{x+y} - \frac{1}{x-y} = \frac{(x-y) - (x+y)}{(x+y)(x-y)} = \frac{x-y-x-y}{(x+y)(x-y)} = \frac{-2y}{(x+y)(x-y)} = \frac{-2y}{x^2 - y^2}x+y1−x−y1=(x+y)(x−y)(x−y)−(x+y)=(x+y)(x−y)x−y−x−y=(x+y)(x−y)−2y=x2−y2−2y分母:1x+y+1x−y=(x−y)+(x+y)(x+y)(x−y)=x−y+x+y(x+y)(x−y)=2x(x+y)(x−y)=2xx2−y2\frac{1}{x+y} + \frac{1}{x-y} = \frac{(x-y) + (x+y)}{(x+y)(x-y)} = \frac{x-y+x+y}{(x+y)(x-y)} = \frac{2x}{(x+y)(x-y)} = \frac{2x}{x^2 - y^2}x+y1+x−y1=(x+y)(x−y)(x−y)+(x+y)=(x+y)(x−y)x−y+x+y=(x+y)(x−y)2x=x2−y22xしたがって、元の式は次のようになります。−2yx2−y22xx2−y2=−2yx2−y2⋅x2−y22x=−2y2x=−yx\frac{\frac{-2y}{x^2 - y^2}}{\frac{2x}{x^2 - y^2}} = \frac{-2y}{x^2 - y^2} \cdot \frac{x^2 - y^2}{2x} = \frac{-2y}{2x} = -\frac{y}{x}x2−y22xx2−y2−2y=x2−y2−2y⋅2xx2−y2=2x−2y=−xy3. 最終的な答え−yx-\frac{y}{x}−xy