Giải bài 34 trang 50 – SGK Toán lớp 8 tập 1

a) $\dfrac{4x + 13}{5x(x - 7)} - \dfrac{x - 48}{5x (7 - x)}$

$= \dfrac{4x + 13}{5x(x - 7)} - \dfrac{x - 48}{-5x (x - 7)}$

$= \dfrac{4x + 13}{5x(x - 7)} + \dfrac{x - 48}{5x (x - 7)}$

$= \dfrac{4x + 13 + x - 48}{5x(x - 7)}$

$= \dfrac{5x - 35}{5x(x - 7)}$

$= \dfrac{5(x - 7)}{5x(x - 7)}$

$= \dfrac{1}{x}$

b) $\dfrac{1}{x - 5x^2} - \dfrac{25x - 15}{25x^2 - 1}$

$= \dfrac{-1}{5x^2 - x} - \dfrac{25x - 15}{25x^2 - 1}$

$= \dfrac{-1}{x(5x - 1)} - \dfrac{25x - 15}{(5x - 1)(5x + 1)}$

$= \dfrac{-(5x + 1)}{x(5x - 1)(5x + 1)} - \dfrac{x(25x - 15)}{x(5x - 1)(5x + 1)}$

$= \dfrac{-5x - 1}{x(5x - 1)(5x + 1)} - \dfrac{25x^2 - 15x}{x(5x - 1)(5x + 1)}$

$= \dfrac{(-5x - 1) - (25x^2 - 15x)}{x(5x - 1)(5x + 1)}$

$= \dfrac{-5x - 1 - 25x^2 + 15x}{x(5x - 1)(5x + 1)}$

$= \dfrac{-25x^2 + 10x - 1}{x(5x - 1)(5x + 1)}$

$= \dfrac{-(25x^2 - 10x + 1)}{x(5x - 1)(5x + 1)}$

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

$= \dfrac{-(5x - 1)}{x(5x + 1)}$

$= \dfrac{1 - 5x}{x(5x + 1)}$

Lưu ý:  $\dfrac{-A}{B} = \dfrac{A}{-B}$