Solve k-1/k+1/2k^2/1-frac{k-1{2k^3}}

$$\frac { \frac { k - 1 } { k } + \frac { 1 } { 2 k ^ { 2 } } } { 1 - \frac { k - 1 } { 2 k ^ { 3 } } }$$

Evaluate

$$\frac{k}{k+1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of k and 2k^{2} is 2k^{2}. Multiply \frac{k-1}{k} times \frac{2k}{2k}.

$$\frac{\frac{\left(k-1\right)\times 2k}{2k^{2}}+\frac{1}{2k^{2}}}{1-\frac{k-1}{2k^{3}}}$$

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

$$\frac{\frac{\left(k-1\right)\times 2k+1}{2k^{2}}}{1-\frac{k-1}{2k^{3}}}$$

Do the multiplications in \left(k-1\right)\times 2k+1.

$$\frac{\frac{2k^{2}-2k+1}{2k^{2}}}{1-\frac{k-1}{2k^{3}}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2k^{3}}{2k^{3}}.

$$\frac{\frac{2k^{2}-2k+1}{2k^{2}}}{\frac{2k^{3}}{2k^{3}}-\frac{k-1}{2k^{3}}}$$

Since \frac{2k^{3}}{2k^{3}} and \frac{k-1}{2k^{3}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{2k^{2}-2k+1}{2k^{2}}}{\frac{2k^{3}-\left(k-1\right)}{2k^{3}}}$$

Do the multiplications in 2k^{3}-\left(k-1\right).

$$\frac{\frac{2k^{2}-2k+1}{2k^{2}}}{\frac{2k^{3}-k+1}{2k^{3}}}$$

Divide \frac{2k^{2}-2k+1}{2k^{2}} by \frac{2k^{3}-k+1}{2k^{3}} by multiplying \frac{2k^{2}-2k+1}{2k^{2}} by the reciprocal of \frac{2k^{3}-k+1}{2k^{3}}.

$$\frac{\left(2k^{2}-2k+1\right)\times 2k^{3}}{2k^{2}\left(2k^{3}-k+1\right)}$$

Cancel out 2k^{2} in both numerator and denominator.

$$\frac{k\left(2k^{2}-2k+1\right)}{2k^{3}-k+1}$$

Factor the expressions that are not already factored.

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

Cancel out 2k^{2}-2k+1 in both numerator and denominator.

$$\frac{k}{k+1}$$

Expand

$$\frac{k}{k+1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of k and 2k^{2} is 2k^{2}. Multiply \frac{k-1}{k} times \frac{2k}{2k}.

$$\frac{\frac{\left(k-1\right)\times 2k}{2k^{2}}+\frac{1}{2k^{2}}}{1-\frac{k-1}{2k^{3}}}$$

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

$$\frac{\frac{\left(k-1\right)\times 2k+1}{2k^{2}}}{1-\frac{k-1}{2k^{3}}}$$

Do the multiplications in \left(k-1\right)\times 2k+1.

$$\frac{\frac{2k^{2}-2k+1}{2k^{2}}}{1-\frac{k-1}{2k^{3}}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2k^{3}}{2k^{3}}.

$$\frac{\frac{2k^{2}-2k+1}{2k^{2}}}{\frac{2k^{3}}{2k^{3}}-\frac{k-1}{2k^{3}}}$$

Since \frac{2k^{3}}{2k^{3}} and \frac{k-1}{2k^{3}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{2k^{2}-2k+1}{2k^{2}}}{\frac{2k^{3}-\left(k-1\right)}{2k^{3}}}$$

Do the multiplications in 2k^{3}-\left(k-1\right).

$$\frac{\frac{2k^{2}-2k+1}{2k^{2}}}{\frac{2k^{3}-k+1}{2k^{3}}}$$

Divide \frac{2k^{2}-2k+1}{2k^{2}} by \frac{2k^{3}-k+1}{2k^{3}} by multiplying \frac{2k^{2}-2k+1}{2k^{2}} by the reciprocal of \frac{2k^{3}-k+1}{2k^{3}}.

$$\frac{\left(2k^{2}-2k+1\right)\times 2k^{3}}{2k^{2}\left(2k^{3}-k+1\right)}$$

Cancel out 2k^{2} in both numerator and denominator.

$$\frac{k\left(2k^{2}-2k+1\right)}{2k^{3}-k+1}$$

Factor the expressions that are not already factored.

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

Cancel out 2k^{2}-2k+1 in both numerator and denominator.

$$\frac{k}{k+1}$$