Question

Tính

a) \(\dfrac{x}{x-3}+\dfrac{-9}{x^2-3x}\)

b) \(\dfrac{x-5}{x^2-4x+4}:\dfrac{x^2-25}{2x-4}\)

Answer 1

\(\dfrac{x}{x-3}+\dfrac{-9}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{-9}{x\left(x-3\right)}=\dfrac{x^2-9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}=\dfrac{x+3}{x}\)

\(\dfrac{x-5}{x^2-4x+4}:\dfrac{x^2-25}{2x-4}=\dfrac{x-5}{\left(x-2\right)^2}.\dfrac{2\left(x-2\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{2}{\left(x-2\right)\left(x+5\right)}\)