Giải bài 22 trang 46 – SGK Toán lớp 8 tập 1

Hướng dẫn: 

$A = -(-A),$ ví dụ $(x - 2) = -[-(x - 2)] = -(2 - x)$

Bài giải

a) $\dfrac{2x^2 - x}{x - 1} + \dfrac{x + 1}{1 - x} + \dfrac{2 - x^2}{x - 1}$

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

$= \dfrac{(2x^2 - x) + [-(x + 1)] + (2 - x^2)}{x - 1}$

$= \dfrac{2x^2 - x - x - 1 + 2 - x^2}{x - 1}$

$= \dfrac{x^2 - 2x + 1 }{x - 1}$

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

b) $\dfrac{4 - x^2}{x - 3} + \dfrac{2x - 2x^2}{3 - x} + \dfrac{5 - 4x}{x - 3}$

$= \dfrac{4 - x^2}{x - 3} + \dfrac{-(2x - 2x^2)}{x - 3} + \dfrac{5 - 4x}{x - 3}$

$= \dfrac{(4 - x^2) + [-(2x - 2x^2)] + (5 - 4x)}{x - 3}$

$= \dfrac{4 - x^2 - 2x + 2x^2 + 5 - 4x}{x - 3}$

$= \dfrac{x^2 - 6x + 9}{x - 3}$

$= \dfrac{(x - 3)^2}{x - 3} = x - 3$