Question

$$\frac{(m+n)^{2}}{m^{2}-n^{2}}+\frac{m^{2}+mn}{n^{2}-mn}$$

Answer

-1-m/n

Solution


Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[\frac{{(m+n)}^{2}}{(m+n)(m-n)}+\frac{{m}^{2}+mn}{{n}^{2}-mn}\]
Factor out the common term \(m\).
\[\frac{{(m+n)}^{2}}{(m+n)(m-n)}+\frac{m(m+n)}{{n}^{2}-mn}\]
Factor out the common term \(n\).
\[\frac{{(m+n)}^{2}}{(m+n)(m-n)}+\frac{m(m+n)}{n(n-m)}\]
Simplify  \(\frac{{(m+n)}^{2}}{(m+n)(m-n)}\)  to  \(\frac{m+n}{m-n}\).
\[\frac{m+n}{m-n}+\frac{m(m+n)}{n(n-m)}\]
Rewrite the expression with a common denominator.
\[\frac{(m+n)n(n-m)+m(m+n)(m-n)}{(m-n)n(n-m)}\]
Factor out the common term \(m+n\).
\[\frac{(m+n)(n(n-m)+m(m-n))}{(m-n)n(n-m)}\]
Expand.
\[\frac{(m+n)({n}^{2}-nm+{m}^{2}-mn)}{(m-n)n(n-m)}\]
Collect like terms.
\[\frac{(m+n)({n}^{2}+(-mn-mn)+{m}^{2})}{(m-n)n(n-m)}\]
Simplify  \({n}^{2}+(-mn-mn)+{m}^{2}\)  to  \({n}^{2}-2mn+{m}^{2}\).
\[\frac{(m+n)({n}^{2}-2mn+{m}^{2})}{(m-n)n(n-m)}\]
Rewrite \({n}^{2}-2mn+{m}^{2}\) in the form \({a}^{2}-2ab+{b}^{2}\), where \(a=n\) and \(b=m\).
\[\frac{(m+n)({n}^{2}-2(n)(m)+{m}^{2})}{(m-n)n(n-m)}\]
Use Square of Difference: \({(a-b)}^{2}={a}^{2}-2ab+{b}^{2}\).
\[\frac{(m+n){(n-m)}^{2}}{(m-n)n(n-m)}\]
Simplify.
\[\frac{(m+n)(n-m)}{(m-n)n}\]
Factor out the negative sign in \(n-m\).
\[-\frac{(m+n)(-n+m)}{(m-n)n}\]
Regroup terms.
\[-\frac{(m+n)(m-n)}{(m-n)n}\]
Cancel \(m-n\).
\[-\frac{m+n}{n}\]
Simplify  \(\frac{m+n}{n}\)  to  \(1+\frac{m}{n}\).
\[-(1+\frac{m}{n})\]
Remove parentheses.
\[-1-\frac{m}{n}\]