次の計算をしなさい。 (1) $\frac{1}{x-1} + \frac{1}{x+2}$ (2) $\frac{1}{3x} - \frac{1}{3x+1}$ (3) $\frac{1}{x-1} - \frac{1}{x(x-1)}$ (4) $\frac{2}{x+1} + \frac{3}{x^2+x}$ (5) $\frac{x}{x+1} + \frac{3x-1}{x^2-2x-3}$ (6) $\frac{x+4}{x^2-2x} - \frac{3}{x^2-3x+2}$

1. 問題の内容

次の計算をしなさい。

(1) $\frac{1}{x-1} + \frac{1}{x+2}$

(2) $\frac{1}{3x} - \frac{1}{3x+1}$

(3) $\frac{1}{x-1} - \frac{1}{x(x-1)}$

(4) $\frac{2}{x+1} + \frac{3}{x^2+x}$

(5) $\frac{x}{x+1} + \frac{3x-1}{x^2-2x-3}$

(6) $\frac{x+4}{x^2-2x} - \frac{3}{x^2-3x+2}$

2. 解き方の手順

(1) $\frac{1}{x-1} + \frac{1}{x+2} = \frac{x+2}{(x-1)(x+2)} + \frac{x-1}{(x-1)(x+2)} = \frac{x+2+x-1}{(x-1)(x+2)} = \frac{2x+1}{(x-1)(x+2)} = \frac{2x+1}{x^2+x-2}$

(2) $\frac{1}{3x} - \frac{1}{3x+1} = \frac{3x+1}{3x(3x+1)} - \frac{3x}{3x(3x+1)} = \frac{3x+1-3x}{3x(3x+1)} = \frac{1}{3x(3x+1)} = \frac{1}{9x^2+3x}$

(3) $\frac{1}{x-1} - \frac{1}{x(x-1)} = \frac{x}{x(x-1)} - \frac{1}{x(x-1)} = \frac{x-1}{x(x-1)} = \frac{1}{x}$

(4) $\frac{2}{x+1} + \frac{3}{x^2+x} = \frac{2}{x+1} + \frac{3}{x(x+1)} = \frac{2x}{x(x+1)} + \frac{3}{x(x+1)} = \frac{2x+3}{x(x+1)} = \frac{2x+3}{x^2+x}$

(5) $\frac{x}{x+1} + \frac{3x-1}{x^2-2x-3} = \frac{x}{x+1} + \frac{3x-1}{(x-3)(x+1)} = \frac{x(x-3)}{(x+1)(x-3)} + \frac{3x-1}{(x-3)(x+1)} = \frac{x^2-3x+3x-1}{(x+1)(x-3)} = \frac{x^2-1}{(x+1)(x-3)} = \frac{(x+1)(x-1)}{(x+1)(x-3)} = \frac{x-1}{x-3}$

(6) $\frac{x+4}{x^2-2x} - \frac{3}{x^2-3x+2} = \frac{x+4}{x(x-2)} - \frac{3}{(x-1)(x-2)} = \frac{(x+4)(x-1)}{x(x-2)(x-1)} - \frac{3x}{x(x-1)(x-2)} = \frac{x^2+3x-4-3x}{x(x-1)(x-2)} = \frac{x^2-4}{x(x-1)(x-2)} = \frac{(x-2)(x+2)}{x(x-1)(x-2)} = \frac{x+2}{x(x-1)} = \frac{x+2}{x^2-x}$

3. 最終的な答え

(1) $\frac{2x+1}{x^2+x-2}$

(2) $\frac{1}{9x^2+3x}$

(3) $\frac{1}{x}$

(4) $\frac{2x+3}{x^2+x}$

(5) $\frac{x-1}{x-3}$

(6) $\frac{x+2}{x^2-x}$