Question

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

Answer 1

a ) \(\left(\dfrac{20x}{3y^2}\right):\left(\dfrac{4x^3}{5y}\right)=\dfrac{20x}{3y^2}.\dfrac{5y}{4x^3}=\dfrac{100xy}{12x^3y^2}=\dfrac{25}{3x^2y}\)

b ) Đ/k : \(x\ne-4\)

Ta có : \(\dfrac{4x+12}{\left(x+4\right)^2}:\dfrac{3\left(x+3\right)}{x+4}\)

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

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

\(=\dfrac{4}{3\left(x+4\right)}\)

\(=\dfrac{4}{3x+12}\)