Solve n-n^2/n-m/frac{m^2{n^2-m^2}+1}=

$$\frac { n - \frac { n ^ { 2 } } { n - m } } { \frac { m ^ { 2 } } { n ^ { 2 } - m ^ { 2 } } + 1 } =$$

Evaluate

$$-\frac{m\left(m+n\right)}{n}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply n times \frac{n-m}{n-m}.

$$\frac{\frac{n\left(n-m\right)}{n-m}-\frac{n^{2}}{n-m}}{\frac{m^{2}}{n^{2}-m^{2}}+1}$$

Since \frac{n\left(n-m\right)}{n-m} and \frac{n^{2}}{n-m} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{n\left(n-m\right)-n^{2}}{n-m}}{\frac{m^{2}}{n^{2}-m^{2}}+1}$$

Do the multiplications in n\left(n-m\right)-n^{2}.

$$\frac{\frac{n^{2}-nm-n^{2}}{n-m}}{\frac{m^{2}}{n^{2}-m^{2}}+1}$$

Combine like terms in n^{2}-nm-n^{2}.

$$\frac{\frac{-nm}{n-m}}{\frac{m^{2}}{n^{2}-m^{2}}+1}$$

Factor n^{2}-m^{2}.

$$\frac{\frac{-nm}{n-m}}{\frac{m^{2}}{\left(m+n\right)\left(-m+n\right)}+1}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\left(m+n\right)\left(-m+n\right)}{\left(m+n\right)\left(-m+n\right)}.

$$\frac{\frac{-nm}{n-m}}{\frac{m^{2}}{\left(m+n\right)\left(-m+n\right)}+\frac{\left(m+n\right)\left(-m+n\right)}{\left(m+n\right)\left(-m+n\right)}}$$

Since \frac{m^{2}}{\left(m+n\right)\left(-m+n\right)} and \frac{\left(m+n\right)\left(-m+n\right)}{\left(m+n\right)\left(-m+n\right)} have the same denominator, add them by adding their numerators.

$$\frac{\frac{-nm}{n-m}}{\frac{m^{2}+\left(m+n\right)\left(-m+n\right)}{\left(m+n\right)\left(-m+n\right)}}$$

Do the multiplications in m^{2}+\left(m+n\right)\left(-m+n\right).

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

Combine like terms in m^{2}-m^{2}+mn-nm+n^{2}.

$$\frac{\frac{-nm}{n-m}}{\frac{n^{2}}{\left(m+n\right)\left(-m+n\right)}}$$

Divide \frac{-nm}{n-m} by \frac{n^{2}}{\left(m+n\right)\left(-m+n\right)} by multiplying \frac{-nm}{n-m} by the reciprocal of \frac{n^{2}}{\left(m+n\right)\left(-m+n\right)}.

$$\frac{-nm\left(m+n\right)\left(-m+n\right)}{\left(n-m\right)n^{2}}$$

Cancel out n\left(-m+n\right) in both numerator and denominator.

$$\frac{-m\left(m+n\right)}{n}$$

Use the distributive property to multiply -m by m+n.

$$\frac{-m^{2}-mn}{n}$$

Expand

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply n times \frac{n-m}{n-m}.

$$\frac{\frac{n\left(n-m\right)}{n-m}-\frac{n^{2}}{n-m}}{\frac{m^{2}}{n^{2}-m^{2}}+1}$$

Since \frac{n\left(n-m\right)}{n-m} and \frac{n^{2}}{n-m} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{n\left(n-m\right)-n^{2}}{n-m}}{\frac{m^{2}}{n^{2}-m^{2}}+1}$$

Do the multiplications in n\left(n-m\right)-n^{2}.

$$\frac{\frac{n^{2}-nm-n^{2}}{n-m}}{\frac{m^{2}}{n^{2}-m^{2}}+1}$$

Combine like terms in n^{2}-nm-n^{2}.

$$\frac{\frac{-nm}{n-m}}{\frac{m^{2}}{n^{2}-m^{2}}+1}$$

Factor n^{2}-m^{2}.

$$\frac{\frac{-nm}{n-m}}{\frac{m^{2}}{\left(m+n\right)\left(-m+n\right)}+1}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\left(m+n\right)\left(-m+n\right)}{\left(m+n\right)\left(-m+n\right)}.

$$\frac{\frac{-nm}{n-m}}{\frac{m^{2}}{\left(m+n\right)\left(-m+n\right)}+\frac{\left(m+n\right)\left(-m+n\right)}{\left(m+n\right)\left(-m+n\right)}}$$

Since \frac{m^{2}}{\left(m+n\right)\left(-m+n\right)} and \frac{\left(m+n\right)\left(-m+n\right)}{\left(m+n\right)\left(-m+n\right)} have the same denominator, add them by adding their numerators.

$$\frac{\frac{-nm}{n-m}}{\frac{m^{2}+\left(m+n\right)\left(-m+n\right)}{\left(m+n\right)\left(-m+n\right)}}$$

Do the multiplications in m^{2}+\left(m+n\right)\left(-m+n\right).

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

Combine like terms in m^{2}-m^{2}+mn-nm+n^{2}.

$$\frac{\frac{-nm}{n-m}}{\frac{n^{2}}{\left(m+n\right)\left(-m+n\right)}}$$

Divide \frac{-nm}{n-m} by \frac{n^{2}}{\left(m+n\right)\left(-m+n\right)} by multiplying \frac{-nm}{n-m} by the reciprocal of \frac{n^{2}}{\left(m+n\right)\left(-m+n\right)}.

$$\frac{-nm\left(m+n\right)\left(-m+n\right)}{\left(n-m\right)n^{2}}$$

Cancel out n\left(-m+n\right) in both numerator and denominator.

$$\frac{-m\left(m+n\right)}{n}$$

Use the distributive property to multiply -m by m+n.

$$\frac{-m^{2}-mn}{n}$$