Giải bài 50 trang 58 – SGK Toán lớp 8 tập 1

a) $\left( \dfrac{x}{x + 1} + 1 \right) : \left(1 - \dfrac{3x^2}{1 - x^2} \right)$

$= \left( \dfrac{x}{x + 1} + \dfrac{x + 1}{x + 1} \right) : \left(\dfrac{1 - x^2}{1 - x^2} - \dfrac{3x^2}{1 - x^2} \right)$

$= \dfrac{x + x + 1}{x + 1} : \dfrac{1 - x^2 - 3x^2}{1 - x^2}$

$= \dfrac{2x + 1}{x + 1} : \dfrac{1 - 4x^2 }{1 - x^2}$

$= \dfrac{2x + 1}{x + 1} : \dfrac{-(4x^2 - 1) }{-(x^2 - 1)}$

$= \dfrac{2x + 1}{x + 1} : \dfrac{4x^2 - 1 }{x^2 - 1}$

$= \dfrac{2x + 1}{x + 1}. \dfrac{ x^2 - 1}{4x^2 - 1}$

$= \dfrac{(2x + 1)(x^2 - 1)}{(x + 1)(4x^2 - 1)}$

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

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

b)  $(x^2 - 1)\left(\dfrac{1}{x - 1} - \dfrac{1}{x + 1} - 1 \right)$

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

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

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

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

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

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

$= -x^2 + 3$

Lưu ý:  $\dfrac{A}{B}:\dfrac{C}{D} = \dfrac{A}{B}.\dfrac{D}{C}$