Solve k^2+1/k-1-k/frac{k^2-1{k+1}+1}*(-2/1+frac{1{k}})=

$$\frac { \frac { k ^ { 2 } + 1 } { k - 1 } - k } { \frac { k ^ { 2 } - 1 } { k + 1 } + 1 } \cdot ( - \frac { 2 } { 1 + \frac { 1 } { k } } ) =$$

Evaluate

$$-\frac{2}{k-1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply k times \frac{k-1}{k-1}.

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

Since \frac{k^{2}+1}{k-1} and \frac{k\left(k-1\right)}{k-1} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in k^{2}+1-k\left(k-1\right).

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

Combine like terms in k^{2}+1-k^{2}+k.

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

Factor the expressions that are not already factored in \frac{k^{2}-1}{k+1}.

$$\frac{\frac{1+k}{k-1}}{\frac{\left(k-1\right)\left(k+1\right)}{k+1}+1}\left(-\frac{2}{1+\frac{1}{k}}\right)$$

Cancel out k+1 in both numerator and denominator.

$$\frac{\frac{1+k}{k-1}}{k-1+1}\left(-\frac{2}{1+\frac{1}{k}}\right)$$

Add -1 and 1 to get 0.

$$\frac{\frac{1+k}{k-1}}{k}\left(-\frac{2}{1+\frac{1}{k}}\right)$$

Express \frac{\frac{1+k}{k-1}}{k} as a single fraction.

$$\frac{1+k}{\left(k-1\right)k}\left(-\frac{2}{1+\frac{1}{k}}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{k}{k}.

$$\frac{1+k}{\left(k-1\right)k}\left(-\frac{2}{\frac{k}{k}+\frac{1}{k}}\right)$$

Since \frac{k}{k} and \frac{1}{k} have the same denominator, add them by adding their numerators.

$$\frac{1+k}{\left(k-1\right)k}\left(-\frac{2}{\frac{k+1}{k}}\right)$$

Divide 2 by \frac{k+1}{k} by multiplying 2 by the reciprocal of \frac{k+1}{k}.

$$\frac{1+k}{\left(k-1\right)k}\left(-\frac{2k}{k+1}\right)$$

Multiply \frac{1+k}{\left(k-1\right)k} times -\frac{2k}{k+1} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-\left(1+k\right)\times 2k}{\left(k-1\right)k\left(k+1\right)}$$

Cancel out k\left(k+1\right) in both numerator and denominator.

$$\frac{-2}{k-1}$$

Expand

$$-\frac{2}{k-1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply k times \frac{k-1}{k-1}.

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

Since \frac{k^{2}+1}{k-1} and \frac{k\left(k-1\right)}{k-1} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in k^{2}+1-k\left(k-1\right).

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

Combine like terms in k^{2}+1-k^{2}+k.

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

Factor the expressions that are not already factored in \frac{k^{2}-1}{k+1}.

$$\frac{\frac{1+k}{k-1}}{\frac{\left(k-1\right)\left(k+1\right)}{k+1}+1}\left(-\frac{2}{1+\frac{1}{k}}\right)$$

Cancel out k+1 in both numerator and denominator.

$$\frac{\frac{1+k}{k-1}}{k-1+1}\left(-\frac{2}{1+\frac{1}{k}}\right)$$

Add -1 and 1 to get 0.

$$\frac{\frac{1+k}{k-1}}{k}\left(-\frac{2}{1+\frac{1}{k}}\right)$$

Express \frac{\frac{1+k}{k-1}}{k} as a single fraction.

$$\frac{1+k}{\left(k-1\right)k}\left(-\frac{2}{1+\frac{1}{k}}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{k}{k}.

$$\frac{1+k}{\left(k-1\right)k}\left(-\frac{2}{\frac{k}{k}+\frac{1}{k}}\right)$$

Since \frac{k}{k} and \frac{1}{k} have the same denominator, add them by adding their numerators.

$$\frac{1+k}{\left(k-1\right)k}\left(-\frac{2}{\frac{k+1}{k}}\right)$$

Divide 2 by \frac{k+1}{k} by multiplying 2 by the reciprocal of \frac{k+1}{k}.

$$\frac{1+k}{\left(k-1\right)k}\left(-\frac{2k}{k+1}\right)$$

Multiply \frac{1+k}{\left(k-1\right)k} times -\frac{2k}{k+1} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-\left(1+k\right)\times 2k}{\left(k-1\right)k\left(k+1\right)}$$

Cancel out k\left(k+1\right) in both numerator and denominator.

$$\frac{-2}{k-1}$$