\(\frac{1}{x+2};\frac{8}{2x-x^2}\)
MTC = x(x + 2)(x - 2)
\(\frac{1}{x+2}=\frac{1.x.\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}=\frac{x\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}\)
\(\frac{8}{2x-x^2}=\frac{-8}{x^2-2x}=\frac{-8}{x\left(x-2\right)}=\frac{-8\left(x+2\right)}{x\left(x+2\right)\left(x-2\right)}\)