Will mark as brainliest :) Simplify the Difference.

Answer:[tex] \frac{-2x \times (x+3)}{(x-2)(x+1)(x-1)}[/tex]Step-by-step explanation:we are given [tex] \frac{2x}{x^2-x-2} - \frac{4x}{x^2-3x+2}[/tex]           ----------(A)Let us simplify the two denominators first. One by one[tex]x^2-x-2[/tex]= [tex]x^2-2x+x-2[/tex]= [tex]x(x-2)-1(x-2)[/tex]= [tex](x-2)(x+1)[/tex][tex]{x^2-3x+2}[/tex]=[tex]x^2-x-2x+2[/tex]=[tex]x(x-1)-2(x-1)[/tex]=[tex](x-1)(x-2)[/tex]Hence (A) becomes[tex]\frac{2x}{x^2-x-2} - \frac{4x}{x^2-3x+2}[/tex]= [tex]\frac{2x}{(x-2)(x+1)} - \frac{4x}{(x-2)(x-1)}[/tex]taking out [tex]\frac{2x}{x-2}[/tex]  as GCD[tex]\frac{2x}{x-2}( \frac{1}{x+1} - \frac{2}{x-1}[/tex][tex]\frac{2x}{x-2}(\frac{(x-1)-2(x+1)}{(x+1)(x-1)}[/tex][tex]\frac{2x}{x-2}(\frac{x-1-2x-2)}{(x+1)(x-1)}[/tex][tex]\frac{2x}{x-2}(\frac{(-x-3)}{(x+1)(x-1)}[/tex][tex]\frac{-2x \times (x+3)}{(x-2)(x+1)(x-1)}[/tex]