Question

\(\dfrac{2x+4}{4x^2-x}:\dfrac{x^2+2x}{1-4x}\)

Answer 1

=\(\dfrac{2\left(x+2\right)}{x\left(4x-1\right)}:\dfrac{x\left(x+2\right)}{-\left(4x-1\right)}\)

=\(\dfrac{2\left(x+2\right)}{x\left(4x-1\right)}.\dfrac{-\left(4x-1\right)}{x\left(x+2\right)}\)= \(\dfrac{-2}{x^2}\)

Answer 2

\(\dfrac{2x+4}{4x^2-x}:\dfrac{x^2+2x}{1-4x}=\dfrac{-2x-4}{x-4x^2}.\dfrac{1-4x}{x^2+2x}\)

=\(\dfrac{-2\left(x+2\right)}{x\left(1-4x\right)}.\dfrac{1-4x}{x\left(x+2\right)}=\dfrac{-2}{x^2}\)