Answer
$\dfrac{(m+n)^{2}}{m-n}\cdot\dfrac{m}{m^{2}+mn}=\dfrac{m+n}{m-n}$
Work Step by Step
$\dfrac{(m+n)^{2}}{m-n}\cdot\dfrac{m}{m^{2}+mn}$
Factor the denominator of the second fraction by taking out common factor $m$:
$\dfrac{(m+n)^{2}}{m-n}\cdot\dfrac{m}{m^{2}+mn}=\dfrac{(m+n)^{2}}{m-n}\cdot\dfrac{m}{m(m+n)}=...$
Multipy both rational expressions and simplify by removing repeated factors in the numerator and the denominator:
$...=\dfrac{m(m+n)^{2}}{m(m+n)(m-n)}=\dfrac{(m+n)^{2}}{(m+n)(m-n)}=\dfrac{m+n}{m-n}$
