第1章 1.39 | 高専の数学解説

(1)

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

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

(2)

$\frac{1}{x^2-x-6} + \frac{x}{x^2-2x-8} = \frac{1}{(x-3)(x+2)} + \frac{x}{(x-4)(x+2)} = \frac{x-4+x(x-3)}{(x-3)(x+2)(x-4)} = \frac{x^2-2x-4}{(x-3)(x+2)(x-4)}$

(3)

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

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

(4)

$\frac{x+2}{2x^2-5x+3} + \frac{3x-1}{2x^2+x-6} + \frac{2x^2-5}{x^2+x-2} = \frac{x+2}{(2x-3)(x-1)} + \frac{(3x-1)}{(2x-3)(x+2)} + \frac{(2x^2-5)}{(x+2)(x-1)}$

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

(5)

$\frac{x^2+x-2}{x^2-2x-8} + \frac{x^2-4x-21}{x^2-x-12} - \frac{x^2+3x-18}{x^2+6x} = \frac{(x+2)(x-1)}{(x-4)(x+2)} + \frac{(x-7)(x+3)}{(x-4)(x+3)} - \frac{(x+6)(x-3)}{x(x+6)}$

$= \frac{x-1}{x-4} + \frac{x-7}{x-4} - \frac{x-3}{x} = \frac{x(x-1)+x(x-7)-(x-3)(x-4)}{x(x-4)} = \frac{x^2-x-12}{x(x-4)} = \frac{(x-4)(x+3)}{x(x-4)} = \frac{x+3}{x}$