Solve [(m-n/m+n-m+n/m-n):m+n/mn](1/m-frac{1}{n})+frac{3-4mn}{(m+n)^2}

$$[ ( \frac { m - n } { m + n } - \frac { m + n } { m - n } ) : \frac { m + n } { m n } ] ( \frac { 1 } { m } - \frac { 1 } { n } ) + \frac { 3 - 4 m n } { ( m + n ) ^ { 2 } }$$

Evaluate

$$\frac{3}{\left(m+n\right)^{2}}$$

Short Solution Steps

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

$$\frac{\frac{\left(m-n\right)\left(m-n\right)}{\left(m+n\right)\left(m-n\right)}-\frac{\left(m+n\right)\left(m+n\right)}{\left(m+n\right)\left(m-n\right)}}{\frac{m+n}{mn}}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Since \frac{\left(m-n\right)\left(m-n\right)}{\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, subtract them by subtracting their numerators.

$$\frac{\frac{\left(m-n\right)\left(m-n\right)-\left(m+n\right)\left(m+n\right)}{\left(m+n\right)\left(m-n\right)}}{\frac{m+n}{mn}}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Do the multiplications in \left(m-n\right)\left(m-n\right)-\left(m+n\right)\left(m+n\right).

$$\frac{\frac{m^{2}-mn-mn+n^{2}-m^{2}-mn-mn-n^{2}}{\left(m+n\right)\left(m-n\right)}}{\frac{m+n}{mn}}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

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

$$\frac{\frac{-4mn}{\left(m+n\right)\left(m-n\right)}}{\frac{m+n}{mn}}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Divide \frac{-4mn}{\left(m+n\right)\left(m-n\right)} by \frac{m+n}{mn} by multiplying \frac{-4mn}{\left(m+n\right)\left(m-n\right)} by the reciprocal of \frac{m+n}{mn}.

$$\frac{-4mnmn}{\left(m+n\right)\left(m-n\right)\left(m+n\right)}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Multiply m and m to get m^{2}.

$$\frac{-4m^{2}nn}{\left(m+n\right)\left(m-n\right)\left(m+n\right)}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Multiply n and n to get n^{2}.

$$\frac{-4m^{2}n^{2}}{\left(m+n\right)\left(m-n\right)\left(m+n\right)}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Multiply m+n and m+n to get \left(m+n\right)^{2}.

$$\frac{-4m^{2}n^{2}}{\left(m+n\right)^{2}\left(m-n\right)}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of m and n is mn. Multiply \frac{1}{m} times \frac{n}{n}. Multiply \frac{1}{n} times \frac{m}{m}.

$$\frac{-4m^{2}n^{2}}{\left(m+n\right)^{2}\left(m-n\right)}\left(\frac{n}{mn}-\frac{m}{mn}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Since \frac{n}{mn} and \frac{m}{mn} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-4m^{2}n^{2}}{\left(m+n\right)^{2}\left(m-n\right)}\times \frac{n-m}{mn}+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Multiply \frac{-4m^{2}n^{2}}{\left(m+n\right)^{2}\left(m-n\right)} times \frac{n-m}{mn} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-4m^{2}n^{2}\left(n-m\right)}{\left(m+n\right)^{2}\left(m-n\right)mn}+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Extract the negative sign in n-m.

$$\frac{-4\left(-1\right)\left(m-n\right)m^{2}n^{2}}{mn\left(m-n\right)\left(m+n\right)^{2}}+\frac{3-4mn}{\left(m+n\right)^{2}}$$

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

$$\frac{-4\left(-1\right)mn}{\left(m+n\right)^{2}}+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Since \frac{-4\left(-1\right)mn}{\left(m+n\right)^{2}} and \frac{3-4mn}{\left(m+n\right)^{2}} have the same denominator, add them by adding their numerators.

$$\frac{-4\left(-1\right)mn+3-4mn}{\left(m+n\right)^{2}}$$

Do the multiplications in -4\left(-1\right)mn+3-4mn.

$$\frac{4mn+3-4mn}{\left(m+n\right)^{2}}$$

Combine like terms in 4mn+3-4mn.

$$\frac{3}{\left(m+n\right)^{2}}$$

Expand \left(m+n\right)^{2}.

$$\frac{3}{m^{2}+2mn+n^{2}}$$

Expand

$$\frac{3}{\left(m+n\right)^{2}}$$

Short Solution Steps

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

$$\frac{\frac{\left(m-n\right)\left(m-n\right)}{\left(m+n\right)\left(m-n\right)}-\frac{\left(m+n\right)\left(m+n\right)}{\left(m+n\right)\left(m-n\right)}}{\frac{m+n}{mn}}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Since \frac{\left(m-n\right)\left(m-n\right)}{\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, subtract them by subtracting their numerators.

$$\frac{\frac{\left(m-n\right)\left(m-n\right)-\left(m+n\right)\left(m+n\right)}{\left(m+n\right)\left(m-n\right)}}{\frac{m+n}{mn}}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Do the multiplications in \left(m-n\right)\left(m-n\right)-\left(m+n\right)\left(m+n\right).

$$\frac{\frac{m^{2}-mn-mn+n^{2}-m^{2}-mn-mn-n^{2}}{\left(m+n\right)\left(m-n\right)}}{\frac{m+n}{mn}}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

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

$$\frac{\frac{-4mn}{\left(m+n\right)\left(m-n\right)}}{\frac{m+n}{mn}}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Divide \frac{-4mn}{\left(m+n\right)\left(m-n\right)} by \frac{m+n}{mn} by multiplying \frac{-4mn}{\left(m+n\right)\left(m-n\right)} by the reciprocal of \frac{m+n}{mn}.

$$\frac{-4mnmn}{\left(m+n\right)\left(m-n\right)\left(m+n\right)}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Multiply m and m to get m^{2}.

$$\frac{-4m^{2}nn}{\left(m+n\right)\left(m-n\right)\left(m+n\right)}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Multiply n and n to get n^{2}.

$$\frac{-4m^{2}n^{2}}{\left(m+n\right)\left(m-n\right)\left(m+n\right)}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Multiply m+n and m+n to get \left(m+n\right)^{2}.

$$\frac{-4m^{2}n^{2}}{\left(m+n\right)^{2}\left(m-n\right)}\left(\frac{1}{m}-\frac{1}{n}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of m and n is mn. Multiply \frac{1}{m} times \frac{n}{n}. Multiply \frac{1}{n} times \frac{m}{m}.

$$\frac{-4m^{2}n^{2}}{\left(m+n\right)^{2}\left(m-n\right)}\left(\frac{n}{mn}-\frac{m}{mn}\right)+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Since \frac{n}{mn} and \frac{m}{mn} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-4m^{2}n^{2}}{\left(m+n\right)^{2}\left(m-n\right)}\times \frac{n-m}{mn}+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Multiply \frac{-4m^{2}n^{2}}{\left(m+n\right)^{2}\left(m-n\right)} times \frac{n-m}{mn} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-4m^{2}n^{2}\left(n-m\right)}{\left(m+n\right)^{2}\left(m-n\right)mn}+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Extract the negative sign in n-m.

$$\frac{-4\left(-1\right)\left(m-n\right)m^{2}n^{2}}{mn\left(m-n\right)\left(m+n\right)^{2}}+\frac{3-4mn}{\left(m+n\right)^{2}}$$

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

$$\frac{-4\left(-1\right)mn}{\left(m+n\right)^{2}}+\frac{3-4mn}{\left(m+n\right)^{2}}$$

Since \frac{-4\left(-1\right)mn}{\left(m+n\right)^{2}} and \frac{3-4mn}{\left(m+n\right)^{2}} have the same denominator, add them by adding their numerators.

$$\frac{-4\left(-1\right)mn+3-4mn}{\left(m+n\right)^{2}}$$

Do the multiplications in -4\left(-1\right)mn+3-4mn.

$$\frac{4mn+3-4mn}{\left(m+n\right)^{2}}$$

Combine like terms in 4mn+3-4mn.

$$\frac{3}{\left(m+n\right)^{2}}$$

Expand \left(m+n\right)^{2}.

$$\frac{3}{m^{2}+2mn+n^{2}}$$