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]
