Solve 1+2/p/p+frac{4{p}}

$$\frac { 1 + \frac { 2 } { p } } { p + \frac { 4 } { p } }$$

Evaluate

$$\frac{p+2}{p^{2}+4}$$

Short Solution Steps

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

$$\frac{\frac{p}{p}+\frac{2}{p}}{p+\frac{4}{p}}$$

Since \frac{p}{p} and \frac{2}{p} have the same denominator, add them by adding their numerators.

$$\frac{\frac{p+2}{p}}{p+\frac{4}{p}}$$

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

$$\frac{\frac{p+2}{p}}{\frac{pp}{p}+\frac{4}{p}}$$

Since \frac{pp}{p} and \frac{4}{p} have the same denominator, add them by adding their numerators.

$$\frac{\frac{p+2}{p}}{\frac{pp+4}{p}}$$

Do the multiplications in pp+4.

$$\frac{\frac{p+2}{p}}{\frac{p^{2}+4}{p}}$$

Divide \frac{p+2}{p} by \frac{p^{2}+4}{p} by multiplying \frac{p+2}{p} by the reciprocal of \frac{p^{2}+4}{p}.

$$\frac{\left(p+2\right)p}{p\left(p^{2}+4\right)}$$

Cancel out p in both numerator and denominator.

$$\frac{p+2}{p^{2}+4}$$

Expand

$$\frac{p+2}{p^{2}+4}$$

Short Solution Steps

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

$$\frac{\frac{p}{p}+\frac{2}{p}}{p+\frac{4}{p}}$$

Since \frac{p}{p} and \frac{2}{p} have the same denominator, add them by adding their numerators.

$$\frac{\frac{p+2}{p}}{p+\frac{4}{p}}$$

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

$$\frac{\frac{p+2}{p}}{\frac{pp}{p}+\frac{4}{p}}$$

Since \frac{pp}{p} and \frac{4}{p} have the same denominator, add them by adding their numerators.

$$\frac{\frac{p+2}{p}}{\frac{pp+4}{p}}$$

Do the multiplications in pp+4.

$$\frac{\frac{p+2}{p}}{\frac{p^{2}+4}{p}}$$

Divide \frac{p+2}{p} by \frac{p^{2}+4}{p} by multiplying \frac{p+2}{p} by the reciprocal of \frac{p^{2}+4}{p}.

$$\frac{\left(p+2\right)p}{p\left(p^{2}+4\right)}$$

Cancel out p in both numerator and denominator.

$$\frac{p+2}{p^{2}+4}$$