Answer
$\frac{8x^3-32x^2+23x+6}{(x+2)(x-1)(x-2)(x-2)}$.
Work Step by Step
The given expression is
$=\frac{3x}{x^2-4}+\frac{5x}{x^2+x-2}-\frac{3}{x^2-4x+4}$
First denominator $=x^2-4$.
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=x^2-(2)^2$
$=(x+2)(x-2)$
Second denominator $=x^2+x-2$.
Rewrite the middle term $x$ as $2x-x$
$=x^2+2x-x-2$
Group terms.
$=(x^2+2x)+(-x-2)$
Factor each term.
$=x(x+2)-1(x+2)$
Factor out $(x+2)$.
$=(x+2)(x-1)$.
Second denominator $=x^2-4x+4$.
Rewrite middle term $-4x$ as $-2x-2x$
$=x^2-2x-2x+4$
Group terms.
$=(x^2-2x)+(-2x+4)$
Factor each term.
$=x(x-2)-2(x-2)$
Factor out $(x-2)$.
$=(x-2)(x-2)$
Substitute all factors into the given expression.
$=\frac{3x}{(x+2)(x-2)}+\frac{5x}{(x+2)(x-1)}-\frac{3}{(x-2)(x-2)}$
The LCM of all the denominators is
$=(x+2)(x-1)(x-2)(x-2)$.
Multiply all fractions to make denominators equal.
$=\frac{3x}{(x+2)(x-2)}\times \frac{(x-2)(x-1)}{(x-2)(x-1)}+\frac{5x}{(x+2)(x-1)}\times \frac{(x-2)(x-2)}{(x-2)(x-2)}-\frac{3}{(x-2)(x-2)}\times \frac{(x+2)(x-1)}{(x+2)(x-1)}$
Simplify.
$=\frac{3x(x-2)(x-1)}{(x+2)(x-1)(x-2)(x-2)}+\frac{5x(x-2)(x-2)}{(x+2)(x-1)(x-2)(x-2)}-\frac{3(x+2)(x-1)}{(x+2)(x-1)(x-2)(x-2)}$
$=\frac{3x(x-2)(x-1)+5x(x-2)(x-2)-3(x+2)(x-1)}{(x+2)(x-1)(x-2)(x-2)}$
$=\frac{3x(x^2-x-2x+2)+5x(x^2-4x+4)-3(x^2+2x-x-2)}{(x+2)(x-1)(x-2)(x-2)}$
Simplify.
$=\frac{3x^3-9x^2+6x+5x^3-20x^2+20x-3x^2-3x+6}{(x+2)(x-1)(x-2)(x-2)}$
$=\frac{8x^3-32x^2+23x+6}{(x+2)(x-1)(x-2)(x-2)}$.
