Giải bài 42 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) $\left(-\dfrac{20x}{3y^2}\right) : \left(-\dfrac{4x^3}{5y}\right)$

$= \left(-\dfrac{20x}{3y^2}\right).\left(-\dfrac{5y}{4x^3}\right)$

$= \dfrac{20x.5y}{3y^2.4x^3} = \dfrac{25}{3x^2y}$

b) $\dfrac{4x + 12}{(x + 4)^2}: \dfrac{3(x + 3)}{x + 4}$

$= \dfrac{4(x + 3)}{(x + 4)^2}: \dfrac{3(x + 3)}{x + 4}$

$= \dfrac{4(x + 3)}{(x + 4)^2}. \dfrac{x + 4}{3(x + 3)}$

$= \dfrac{4(x + 3)(x + 4)}{3(x + 4)^2(x + 3)}$

$= \dfrac{4}{3(x + 4)}$