Làm tính chia phân thức:

a) $\left(-\frac{20x}{y^2}\right):\left(\frac{4x^3}{5y}\right);$                                             b) $\frac{4x+12}{{\left(x+4\right)}^2}:\frac{3(x+3)}{x+4}$.

Giải

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

Áp dụng công thức:

$\left(-\frac{A}{B}\right):\left(-\frac{C}{D}\right)=\left(\frac{A}{B}\right):\left(\frac{C}{D}\right)$

Ta được: $\left(-\frac{20x}{y^2}\right):\left(\frac{4x^3}{5y}\right)=\frac{20x}{y^2}:\frac{4x^3}{5y}=\frac{20x}{y^2}.\frac{5y}{4x^3}=\frac{20x.5y}{y^2.4x^3}=\frac{25xy}{4x^3{y}^2}=\frac{25}{3x^{2}y}.$

b) $\frac{4x+12}{{\left(x+4\right)}^2}:\frac{3(x+3)}{x+4}=\frac{4x+12}{{\left(x+4\right)}^2}.\frac{x+4}{3(x+3)}=\frac{4}{3\left(x+4\right)}$