=>\(\frac{x^2-3^2-\left(4x-2\right)\cdot\left(x-3\right)}{\left(x-3\right)^2}\)
=>\(\frac{\left(x+3\right)\cdot\left(x-3\right)-\left(4x-2\right)\cdot\left(x-3\right)}{\left(x-3\right)^2}\)
=>\(\frac{\left(x-3\right)\cdot\left(x+3-4x+2\right)}{\left(x-3\right)^2}\)
=>\(\frac{-3x+5}{x-3}\)
cho minh nhe!