Simplify: (m²−n²)/(m−2mn+n²) ÷ (m²+mn)/(n²-mn)

Question

ScanSolve AI Tutor · Accepted Answer

To simplify the given expression, we will factorize each polynomial and then perform the division.

The expression is:

\frac{m^2 - n^2}{m^2 - 2mn + n^2} \div \frac{m^2 + mn}{n^2 - mn}

Factorize each part of the fractions.
- Numerator of the first fraction: $m^2 - n^2$ This is a difference of squares, which factors as $(m-n)(m+n)$ .
- Denominator of the first fraction: $m^2 - 2mn + n^2$ This is a perfect square trinomial, which factors as $(m-n)^2$ .
- Numerator of the second fraction: $m^2 + mn$ Factor out the common term $m$ : $m(m+n)$ .
- Denominator of the second fraction: $n^2 - mn$ Factor out the common term $n$ : $n(n-m)$ . We can rewrite $(n-m)$ as $-(m-n)$ , so $n(n-m) = -n(m-n)$ .
Substitute the factored forms back into the expression.
$\frac{(m-n)(m+n)}{(m-n)^2} \div \frac{m(m+n)}{-n(m-n)}$
Change the division to multiplication by taking the reciprocal of the second fraction.
$\frac{(m-n)(m+n)}{(m-n)^2} \times \frac{-n(m-n)}{m(m+n)}$
Cancel out common factors from the numerator and denominator.
$\frac{\cancel{(m-n)}(m+n)}{(m-n)\cancel{(m-n)}} \times \frac{-n\cancel{(m-n)}}{m(m+n)}$ $\frac{m+n}{m-n} \times \frac{-n}{m(m+n)}$
Now, cancel $(m+n)$ from the numerator of the first term and the denominator of the second term.
$\frac{\cancel{(m+n)}}{m-n} \times \frac{-n}{m\cancel{(m+n)}}$
Multiply the remaining terms.
$\frac{1}{m-n} \times \frac{-n}{m} = \frac{-n}{m(m-n)}$
The simplified expression is:
$\frac{-n}{m(m-n)}$

Simplify: (m²-n²)/(m-2mn+n²) ÷ (m²+mn)/(n²-mn)