与えられた分数式の計算をせよ。 $$\frac{2}{x+1} + \frac{2x}{x-1} - \frac{x^2+3}{x^2-1}$$代数学分数式代数計算因数分解式の計算2025/5/61. 問題の内容与えられた分数式の計算をせよ。2x+1+2xx−1−x2+3x2−1\frac{2}{x+1} + \frac{2x}{x-1} - \frac{x^2+3}{x^2-1}x+12+x−12x−x2−1x2+32. 解き方の手順まず、分母を因数分解します。x2−1=(x+1)(x−1)x^2 - 1 = (x+1)(x-1)x2−1=(x+1)(x−1) となります。次に、各分数の分母を (x+1)(x−1)(x+1)(x-1)(x+1)(x−1) に揃えます。2x+1=2(x−1)(x+1)(x−1)=2x−2(x+1)(x−1)\frac{2}{x+1} = \frac{2(x-1)}{(x+1)(x-1)} = \frac{2x-2}{(x+1)(x-1)}x+12=(x+1)(x−1)2(x−1)=(x+1)(x−1)2x−22xx−1=2x(x+1)(x−1)(x+1)=2x2+2x(x+1)(x−1)\frac{2x}{x-1} = \frac{2x(x+1)}{(x-1)(x+1)} = \frac{2x^2+2x}{(x+1)(x-1)}x−12x=(x−1)(x+1)2x(x+1)=(x+1)(x−1)2x2+2xx2+3x2−1=x2+3(x+1)(x−1)\frac{x^2+3}{x^2-1} = \frac{x^2+3}{(x+1)(x-1)}x2−1x2+3=(x+1)(x−1)x2+3分母が揃ったので、分子を計算します。2x−2(x+1)(x−1)+2x2+2x(x+1)(x−1)−x2+3(x+1)(x−1)=2x−2+2x2+2x−(x2+3)(x+1)(x−1)\frac{2x-2}{(x+1)(x-1)} + \frac{2x^2+2x}{(x+1)(x-1)} - \frac{x^2+3}{(x+1)(x-1)} = \frac{2x-2+2x^2+2x-(x^2+3)}{(x+1)(x-1)}(x+1)(x−1)2x−2+(x+1)(x−1)2x2+2x−(x+1)(x−1)x2+3=(x+1)(x−1)2x−2+2x2+2x−(x2+3)=2x−2+2x2+2x−x2−3(x+1)(x−1)=x2+4x−5(x+1)(x−1)= \frac{2x-2+2x^2+2x-x^2-3}{(x+1)(x-1)} = \frac{x^2+4x-5}{(x+1)(x-1)}=(x+1)(x−1)2x−2+2x2+2x−x2−3=(x+1)(x−1)x2+4x−5分子を因数分解します。x2+4x−5=(x+5)(x−1)x^2+4x-5 = (x+5)(x-1)x2+4x−5=(x+5)(x−1) となります。(x+5)(x−1)(x+1)(x−1)=x+5x+1\frac{(x+5)(x-1)}{(x+1)(x-1)} = \frac{x+5}{x+1}(x+1)(x−1)(x+5)(x−1)=x+1x+53. 最終的な答えx+5x+1\frac{x+5}{x+1}x+1x+5