\(\dfrac{4-x}{x^2+2}-\dfrac{x^2+5x}{x^3-x^2+2x-2}\\ =\dfrac{4-x}{x^2+2}-\dfrac{x^2+5x}{x^2\left(x-1\right)+2\left(x-1\right)}\\ =\dfrac{4-x}{x^2+2}-\dfrac{x^2+5x}{\left(x^2+2\right)\left(x-1\right)}\\ =\dfrac{\left(4-x\right)\left(x-1\right)-\left(x^2+5x\right)}{\left(x^2+2\right)\left(x-1\right)}\\ =\dfrac{4x-x^2-4+x-x^2-5x}{\left(x^2+2\right)\left(x-1\right)}\\ =\dfrac{-2x^2-4}{\left(x^2+2\right)\left(x-1\right)}\\ =\dfrac{-2\left(x^2+2\right)}{\left(x^2+2\right)\left(x-1\right)}\\ =\dfrac{-2}{x-1}\)