Answer
$\dfrac{(m-n)^{2}}{m+n}\div\dfrac{m^{2}-mn}{m}=\dfrac{m-n}{m+n}$
Work Step by Step
$\dfrac{(m-n)^{2}}{m+n}\div\dfrac{m^{2}-mn}{m}$
Take out common factor $m$ from the numerator of the second fraction:
$\dfrac{(m-n)^{2}}{m+n}\div\dfrac{m^{2}-mn}{m}=\dfrac{(m-n)^{2}}{m+n}\div\dfrac{m(m-n)}{m}=...$
Evaluate the division of the two rational expressions. Then, simplify by removing repeated factors in the numerator and the denominator:
$...=\dfrac{m(m-n)^{2}}{m(m-n)(m+n)}=\dfrac{m-n}{m+n}$
