与えられた数式を簡略化する問題です。 数式は以下の通りです。 $\frac{1}{x-1} - \frac{2}{x^2+1} - \frac{1}{x+1} - \frac{4}{x^4+1}$代数学分数式式の簡略化代数2025/4/231. 問題の内容与えられた数式を簡略化する問題です。数式は以下の通りです。1x−1−2x2+1−1x+1−4x4+1\frac{1}{x-1} - \frac{2}{x^2+1} - \frac{1}{x+1} - \frac{4}{x^4+1}x−11−x2+12−x+11−x4+142. 解き方の手順まず、1x−1\frac{1}{x-1}x−11 と −1x+1-\frac{1}{x+1}−x+11 をまとめます。1x−1−1x+1=(x+1)−(x−1)(x−1)(x+1)=x+1−x+1x2−1=2x2−1\frac{1}{x-1} - \frac{1}{x+1} = \frac{(x+1) - (x-1)}{(x-1)(x+1)} = \frac{x+1-x+1}{x^2-1} = \frac{2}{x^2-1}x−11−x+11=(x−1)(x+1)(x+1)−(x−1)=x2−1x+1−x+1=x2−12与式は、2x2−1−2x2+1−4x4+1\frac{2}{x^2-1} - \frac{2}{x^2+1} - \frac{4}{x^4+1}x2−12−x2+12−x4+14=2(1x2−1−1x2+1)−4x4+1= 2\left(\frac{1}{x^2-1} - \frac{1}{x^2+1}\right) - \frac{4}{x^4+1}=2(x2−11−x2+11)−x4+14=2((x2+1)−(x2−1)(x2−1)(x2+1))−4x4+1= 2\left(\frac{(x^2+1) - (x^2-1)}{(x^2-1)(x^2+1)}\right) - \frac{4}{x^4+1}=2((x2−1)(x2+1)(x2+1)−(x2−1))−x4+14=2(x2+1−x2+1x4−1)−4x4+1= 2\left(\frac{x^2+1-x^2+1}{x^4-1}\right) - \frac{4}{x^4+1}=2(x4−1x2+1−x2+1)−x4+14=2(2x4−1)−4x4+1= 2\left(\frac{2}{x^4-1}\right) - \frac{4}{x^4+1}=2(x4−12)−x4+14=4x4−1−4x4+1= \frac{4}{x^4-1} - \frac{4}{x^4+1}=x4−14−x4+14=4(1x4−1−1x4+1)= 4\left(\frac{1}{x^4-1} - \frac{1}{x^4+1}\right)=4(x4−11−x4+11)=4((x4+1)−(x4−1)(x4−1)(x4+1))= 4\left(\frac{(x^4+1) - (x^4-1)}{(x^4-1)(x^4+1)}\right)=4((x4−1)(x4+1)(x4+1)−(x4−1))=4(x4+1−x4+1x8−1)= 4\left(\frac{x^4+1-x^4+1}{x^8-1}\right)=4(x8−1x4+1−x4+1)=4(2x8−1)= 4\left(\frac{2}{x^8-1}\right)=4(x8−12)=8x8−1= \frac{8}{x^8-1}=x8−183. 最終的な答え8x8−1\frac{8}{x^8-1}x8−18