Question

Will mark as brainliest :) Simplify the Difference.

Answer 1

Answer:

$\frac{-2x \times (x+3)}{(x-2)(x+1)(x-1)}$

Step-by-step explanation:

we are given

$\frac{2x}{x^2-x-2} - \frac{4x}{x^2-3x+2}$           ----------(A)

Let us simplify the two denominators first. One by one

$x^2-x-2$

= $x^2-2x+x-2$

= $x(x-2)-1(x-2)$

= $(x-2)(x+1)$

${x^2-3x+2}$

=$x^2-x-2x+2$

=$x(x-1)-2(x-1)$

=$(x-1)(x-2)$

Hence (A) becomes

$\frac{2x}{x^2-x-2} - \frac{4x}{x^2-3x+2}$

= $\frac{2x}{(x-2)(x+1)} - \frac{4x}{(x-2)(x-1)}$

taking out $\frac{2x}{x-2}$  as GCD

$\frac{2x}{x-2}( \frac{1}{x+1} - \frac{2}{x-1}$

$\frac{2x}{x-2}(\frac{(x-1)-2(x+1)}{(x+1)(x-1)}$

$\frac{2x}{x-2}(\frac{x-1-2x-2)}{(x+1)(x-1)}$

$\frac{2x}{x-2}(\frac{(-x-3)}{(x+1)(x-1)}$

$\frac{-2x \times (x+3)}{(x-2)(x+1)(x-1)}$