Answer
$\frac{2b+a}{b-2a}$.
Work Step by Step
The given expression is
$=\frac{\frac{2}{a^2}-\frac{1}{ab}-\frac{1}{b^2}}{\frac{1}{a^2}-\frac{3}{ab}+\frac{2}{b^2}}$
Multiply the numerator and the denominator by $a^2b^2$.
$=\frac{a^2b^2}{a^2b^2}\cdot \frac{\frac{2}{a^2}-\frac{1}{ab}-\frac{1}{b^2}}{\frac{1}{a^2}-\frac{3}{ab}+\frac{2}{b^2}}$
Use the distributive property.
$=\frac{a^2b^2\cdot \frac{2}{a^2}-a^2b^2\cdot\frac{1}{ab}-a^2b^2\cdot\frac{1}{b^2}}{a^2b^2\cdot\frac{1}{a^2}-a^2b^2\cdot\frac{3}{ab}+a^2b^2\cdot\frac{2}{b^2}}$
Simplify.
$=\frac{2b^2-ab-a^2}{b^2-3ab+2a^2}$.
Factor the fraction.
Numerator $2b^2-ab-a^2$.
$=2b^2-2ab+ab-a^2$
$=(2b^2-2ab)+(ab-a^2)$
$=2b(b-a)+a(b-a)$
$=(b-a)(2b+a)$
Denominator $b^2-3ab+2a^2$.
$=b^2-3ab+2a^2$
$=b^2-2ab-ab+2a^2$
$=(b^2-2ab)+(-ab+2a^2)$
$=b(b-2a)-a(b-2a)$
$=(b-2a)(b-a)$
Plug all factors into the fraction.
$=\frac{(b-a)(2b+a)}{(b-2a)(b-a)}$.
Cancel common terms.
$=\frac{2b+a}{b-2a}$.
