Solve (2-a/2+a-a+2/a-2):(2+a/2-a+a-2/a+2)

$$( \frac { 2 - a } { 2 + a } - \frac { a + 2 } { a - 2 } ) : ( \frac { 2 + a } { 2 - a } + \frac { a - 2 } { a + 2 } )$$

Evaluate

$$\frac{a}{4}+\frac{1}{a}$$

Short Solution Steps

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

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

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

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

Do the multiplications in \left(2-a\right)\left(a-2\right)-\left(a+2\right)\left(a+2\right).

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

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

$$\frac{\frac{-8-2a^{2}}{\left(a-2\right)\left(a+2\right)}}{\frac{2+a}{2-a}+\frac{a-2}{a+2}}$$

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

$$\frac{\frac{-8-2a^{2}}{\left(a-2\right)\left(a+2\right)}}{\frac{\left(2+a\right)\left(a+2\right)}{\left(a+2\right)\left(-a+2\right)}+\frac{\left(a-2\right)\left(-a+2\right)}{\left(a+2\right)\left(-a+2\right)}}$$

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

$$\frac{\frac{-8-2a^{2}}{\left(a-2\right)\left(a+2\right)}}{\frac{\left(2+a\right)\left(a+2\right)+\left(a-2\right)\left(-a+2\right)}{\left(a+2\right)\left(-a+2\right)}}$$

Do the multiplications in \left(2+a\right)\left(a+2\right)+\left(a-2\right)\left(-a+2\right).

$$\frac{\frac{-8-2a^{2}}{\left(a-2\right)\left(a+2\right)}}{\frac{2a+4+a^{2}+2a-a^{2}+2a+2a-4}{\left(a+2\right)\left(-a+2\right)}}$$

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

$$\frac{\frac{-8-2a^{2}}{\left(a-2\right)\left(a+2\right)}}{\frac{8a}{\left(a+2\right)\left(-a+2\right)}}$$

Divide \frac{-8-2a^{2}}{\left(a-2\right)\left(a+2\right)} by \frac{8a}{\left(a+2\right)\left(-a+2\right)} by multiplying \frac{-8-2a^{2}}{\left(a-2\right)\left(a+2\right)} by the reciprocal of \frac{8a}{\left(a+2\right)\left(-a+2\right)}.

$$\frac{\left(-8-2a^{2}\right)\left(a+2\right)\left(-a+2\right)}{\left(a-2\right)\left(a+2\right)\times 8a}$$

Extract the negative sign in -a+2.

$$\frac{-\left(a-2\right)\left(a+2\right)\left(-2a^{2}-8\right)}{8a\left(a-2\right)\left(a+2\right)}$$

Cancel out \left(a-2\right)\left(a+2\right) in both numerator and denominator.

$$\frac{-\left(-2a^{2}-8\right)}{8a}$$

Factor the expressions that are not already factored.

$$\frac{-2\left(-a^{2}-4\right)}{8a}$$

Cancel out 2 in both numerator and denominator.

$$\frac{-\left(-a^{2}-4\right)}{4a}$$

Expand the expression.

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

Expand

$$\frac{a}{4}+\frac{1}{a}$$

Short Solution Steps

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

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

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

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

Do the multiplications in \left(2-a\right)\left(a-2\right)-\left(a+2\right)\left(a+2\right).

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

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

$$\frac{\frac{-8-2a^{2}}{\left(a-2\right)\left(a+2\right)}}{\frac{2+a}{2-a}+\frac{a-2}{a+2}}$$

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

$$\frac{\frac{-8-2a^{2}}{\left(a-2\right)\left(a+2\right)}}{\frac{\left(2+a\right)\left(a+2\right)}{\left(a+2\right)\left(-a+2\right)}+\frac{\left(a-2\right)\left(-a+2\right)}{\left(a+2\right)\left(-a+2\right)}}$$

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

$$\frac{\frac{-8-2a^{2}}{\left(a-2\right)\left(a+2\right)}}{\frac{\left(2+a\right)\left(a+2\right)+\left(a-2\right)\left(-a+2\right)}{\left(a+2\right)\left(-a+2\right)}}$$

Do the multiplications in \left(2+a\right)\left(a+2\right)+\left(a-2\right)\left(-a+2\right).

$$\frac{\frac{-8-2a^{2}}{\left(a-2\right)\left(a+2\right)}}{\frac{2a+4+a^{2}+2a-a^{2}+2a+2a-4}{\left(a+2\right)\left(-a+2\right)}}$$

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

$$\frac{\frac{-8-2a^{2}}{\left(a-2\right)\left(a+2\right)}}{\frac{8a}{\left(a+2\right)\left(-a+2\right)}}$$

Divide \frac{-8-2a^{2}}{\left(a-2\right)\left(a+2\right)} by \frac{8a}{\left(a+2\right)\left(-a+2\right)} by multiplying \frac{-8-2a^{2}}{\left(a-2\right)\left(a+2\right)} by the reciprocal of \frac{8a}{\left(a+2\right)\left(-a+2\right)}.

$$\frac{\left(-8-2a^{2}\right)\left(a+2\right)\left(-a+2\right)}{\left(a-2\right)\left(a+2\right)\times 8a}$$

Extract the negative sign in -a+2.

$$\frac{-\left(a-2\right)\left(a+2\right)\left(-2a^{2}-8\right)}{8a\left(a-2\right)\left(a+2\right)}$$

Cancel out \left(a-2\right)\left(a+2\right) in both numerator and denominator.

$$\frac{-\left(-2a^{2}-8\right)}{8a}$$

Factor the expressions that are not already factored.

$$\frac{-2\left(-a^{2}-4\right)}{8a}$$

Cancel out 2 in both numerator and denominator.

$$\frac{-\left(-a^{2}-4\right)}{4a}$$

Expand the expression.

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