Giải bài 43 trang 54 – SGK Toán lớp 8 tập 1

Hướng dẫn: $\dfrac{A}{B}: \dfrac{C}{D} = \dfrac{A}{B}.\dfrac{D}{C},$ với $\dfrac{C}{D} \ne 0.$

Bài giải

a) $\dfrac{5x - 10}{x^2 + 7}: (2x - 4)$

$= \dfrac{5x - 10}{x^2 + 7}. \dfrac{1}{2x - 4}$

$= \dfrac{5(x - 2)}{x^2 + 7} . \dfrac{1}{2(x - 2)}$

$= \dfrac{5(x - 2)}{2(x^2 + 7)(x - 2)}$

$= \dfrac{5}{2(x^2 + 7)}$

b) $(x^2 - 25) : \dfrac{2x + 10}{3x - 7}$

$= (x^2 - 25). \dfrac{3x - 7}{2x + 10}$

$= \dfrac{(x^2 - 25)(3x - 7)}{2x + 10}$

$= \dfrac{(x + 5)(x - 5)(3x - 7)}{2(x + 5)}$

$= \dfrac{(x - 5)(3x - 7)}{2}$

c) $\dfrac{x^2 + x}{5x^2 - 10x + 5} : \dfrac{3x + 3}{5x - 5}$

$= \dfrac{x(x + 1)}{5(x^2 - 2x + 1)} : \dfrac{3(x + 1)}{5(x - 1)}$

$= \dfrac{x(x + 1)}{5(x^2 - 2x + 1)}. \dfrac{5(x - 1)}{3(x + 1)}$

$= \dfrac{x(x + 1)}{5(x - 1)^2}. \dfrac{5(x - 1)}{3(x + 1)}$

$= \dfrac{x(x + 1)5(x - 1)}{5(x - 1)^2.3(x + 1)}$

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