Answer
$\frac{m+6}{(m+2)(2m-1)(m-1)}$.
Work Step by Step
The given expression is
$=\frac{1}{m^2+m-2}-\frac{3}{2m^2+3m-2}+\frac{2}{2m^2-3m+1}$
First denominator
$=m^2+m-2$.
Rewrite middle term $m$ as $2m-m$
$=m^2+2m-m-2$.
Group terms.
$=(m^2+2m)+(-m-2)$.
Factor each term.
$=m(m+2)-1(m+2)$.
Factor out $(m+2)$.
$=(m+2)(m-1)$.
Second denominator
$=2m^2+3m-2$.
Rewrite middle term $3m$ as $4m-1m$
$=2m^2+4m-1m-2$.
Group terms.
$=(2m^2+4m)+(-1m-2)$.
Factor each term.
$=2m(m+2)-1(m+2)$.
Factor out $(m+2)$.
$=(m+2)(2m-1)$.
Third denominator
$=2m^2-3m+1$.
Rewrite middle term $-3m$ as $-2m-1m$
$=2m^2-2m-1m+1$.
Group terms.
$=(2m^2-2m)+(-1m+1)$.
Factor each term.
$=2m(m-1)-1(m-1)$.
Factor out $(m-1)$.
$=(m-1)(2m-1)$.
Substitute all factors into the given expression.
$=\frac{1}{(m+2)(m-1)}-\frac{3}{(m+2)(2m-1)}+\frac{2}{(m-1)(2m-1)}$
The LCM of all the denominators is
$=(m+2)(2m-1)(m-1)$.
$=\frac{1}{(m+2)(m-1)}\times \frac{(2m-1)}{(2m-1)}-\frac{3}{(m+2)(2m-1)}\times \frac{(m-1)}{(m-1)}+\frac{2}{(m-1)(2m-1)}\times \frac{(m+2)}{(m+2)}$
$=\frac{(2m-1)}{(m+2)(2m-1)(m-1)}-\frac{3(m-1)}{(m+2)(2m-1)(m-1)}+\frac{2(m+2)}{(m+2)(2m-1)(m-1)}$
$=\frac{(2m-1)-3(m-1)+2(m+2)}{(m+2)(2m-1)(m-1)}$
$=\frac{2m-1-3m+3+2m+4}{(m+2)(2m-1)(m-1)}$
Simplify.
$=\frac{m+6}{(m+2)(2m-1)(m-1)}$.
