Solve 1+h/2k/frac{h^2-1{5k}}

$$\frac { \frac { 1 + h } { 2 k } } { \frac { h ^ { 2 } - 1 } { 5 k } }$$

Evaluate

$$\frac{5}{2\left(h-1\right)}$$

Short Solution Steps

Divide \frac{1+h}{2k} by \frac{h^{2}-1}{5k} by multiplying \frac{1+h}{2k} by the reciprocal of \frac{h^{2}-1}{5k}.

$$\frac{\left(1+h\right)\times 5k}{2k\left(h^{2}-1\right)}$$

Cancel out k in both numerator and denominator.

$$\frac{5\left(h+1\right)}{2\left(h^{2}-1\right)}$$

Factor the expressions that are not already factored.

$$\frac{5\left(h+1\right)}{2\left(h-1\right)\left(h+1\right)}$$

Cancel out h+1 in both numerator and denominator.

$$\frac{5}{2\left(h-1\right)}$$

Expand the expression.

$$\frac{5}{2h-2}$$

Expand

$$\frac{5}{2\left(h-1\right)}$$

Short Solution Steps

Divide \frac{1+h}{2k} by \frac{h^{2}-1}{5k} by multiplying \frac{1+h}{2k} by the reciprocal of \frac{h^{2}-1}{5k}.

$$\frac{\left(1+h\right)\times 5k}{2k\left(h^{2}-1\right)}$$

Cancel out k in both numerator and denominator.

$$\frac{5\left(h+1\right)}{2\left(h^{2}-1\right)}$$

Factor the expressions that are not already factored.

$$\frac{5\left(h+1\right)}{2\left(h-1\right)\left(h+1\right)}$$

Cancel out h+1 in both numerator and denominator.

$$\frac{5}{2\left(h-1\right)}$$

Expand the expression.

$$\frac{5}{2h-2}$$