Câu 1
c) Ta có
$x^3 - 7x - 6 = x^3 + x^2 - x^2 - x - 6x - 6$
$= x^2(x+1) - x(x+1) - 6(x+1)$
$= (x+1)(x^2 - x - 6)$
$= (x+1)[(x^2 - 3x) + (2x - 6)]$
$= (x+1)[x(x-3) + 2(x-3)]$
$= (x+1)(x+2)(x-3)$
Câu 2
a) Ta có
$\dfrac{x+3}{x-6} + \dfrac{x-23}{x-6} + \dfrac{x+2}{x-6} = \dfrac{(x+3) + (x-23) + (x+2)}{x-6}$
$= \dfrac{3x -18}{x-6}$
$= \dfrac{3(x-6)}{x-6} = 3$
b) Ta có
$\dfrac{3}{2x+6} - \dfrac{x-6}{2x^2 + 6x} = \dfrac{3.x}{x(2x+6)} - \dfrac{x-6}{x(2x+6)}$
$= \dfrac{3x-(x-6)}{x(2x+6)}$
$= \dfrac{2x + 6}{x(2x+6)} = x$
