Solve (2n/k+2n-4n^2/k^2+4nk+4n^2)

$$( \frac { 2 n } { k + 2 n } - \frac { 4 n ^ { 2 } } { k ^ { 2 } + 4 n k + 4 n ^ { 2 } } )$$

Evaluate

$$\frac{2kn}{\left(2n+k\right)^{2}}$$

Short Solution Steps

Factor k^{2}+4nk+4n^{2}.

$$\frac{2n}{k+2n}-\frac{4n^{2}}{\left(2n+k\right)^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of k+2n and \left(2n+k\right)^{2} is \left(2n+k\right)^{2}. Multiply \frac{2n}{k+2n} times \frac{2n+k}{2n+k}.

$$\frac{2n\left(2n+k\right)}{\left(2n+k\right)^{2}}-\frac{4n^{2}}{\left(2n+k\right)^{2}}$$

Since \frac{2n\left(2n+k\right)}{\left(2n+k\right)^{2}} and \frac{4n^{2}}{\left(2n+k\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2n\left(2n+k\right)-4n^{2}}{\left(2n+k\right)^{2}}$$

Do the multiplications in 2n\left(2n+k\right)-4n^{2}.

$$\frac{4n^{2}+2nk-4n^{2}}{\left(2n+k\right)^{2}}$$

Combine like terms in 4n^{2}+2nk-4n^{2}.

$$\frac{2nk}{\left(2n+k\right)^{2}}$$

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

$$\frac{2nk}{4n^{2}+4kn+k^{2}}$$

Factor

$$\frac{2kn}{\left(2n+k\right)^{2}}$$