Bài 42 trang 54 SGK Toán 8 tập 1

Làm tính chia phân thức:

a) \(\left( { - \dfrac{{20x}}{{3{y^2}}}} \right):\left( { - \dfrac{{4{x^3}}}{{5y}}} \right)\);

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

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

\(=\dfrac{20x}{3y^{2}}.\dfrac{5y}{4x^{3}}=\dfrac{20x.5y}{3y^{2}.4x^{3}}=\dfrac{25}{3x^{2}y}\)

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

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