Question

Thực hiện phép tính sau:

_{\(\dfrac{2x+6}{3x^2-x}\)}: \(\dfrac{x^2+3x}{1-3x}\)

Answer 1

Ta có:

\(\dfrac{2x+6}{3x^2-x}:\dfrac{x^2+3x}{1-3x}=\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}.\dfrac{1-3x}{x\left(x+3\right)}=-\dfrac{2}{x^2}\)

Answer 2

\(\dfrac{2x+6}{3x^2-x}:\dfrac{x^2+3x}{1-3x}\\ =\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}.\dfrac{-\left(3x-1\right)}{x\left(x+3\right)}\\ =\dfrac{-2}{x^2}\)