Đáp án: + Giải thích các bước giải:
a)
$\dfrac{3}{x² - 3x} + \dfrac{5}{2x - 6}$
$= \dfrac{3}{x(x-3)} + \dfrac{5}{2(x-3)}$
$= \dfrac{3.2}{2x(x-3)} + \dfrac{5.x}{2x(x-3)}$
$= \dfrac{6 + 5x}{2x(x-3)}$
b)
$\dfrac{1}{x+1} + \dfrac{6}{x-x²}$
$= \dfrac{1}{x+1} - \dfrac{6}{x² - x}$
$= \dfrac{1}{x+1} - \dfrac{6}{x(x-1)}$
$= \dfrac{x(x-1)}{(x- 1)(x+1)} - \dfrac{6(x+1)}{x(x+1)(x-1)}$
$= \dfrac{x² - x - 6x - 6}{x(x-1)(x+1)}$
$=\dfrac{x² - 7x - 6}{x(x-1)(x+1)}$
