Solve (x+8/x^2-4-2/x-2)divx-4/x^2-4x+4

$$( \frac { x + 8 } { x ^ { 2 } - 4 } - \frac { 2 } { x - 2 } ) \div \frac { x - 4 } { x ^ { 2 } - 4 x + 4 }$$

Evaluate

$$-\frac{x-2}{x+2}$$

Short Solution Steps

Factor x^{2}-4.

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

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

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

Since \frac{x+8}{\left(x-2\right)\left(x+2\right)} and \frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in x+8-2\left(x+2\right).

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

Combine like terms in x+8-2x-4.

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

Divide \frac{-x+4}{\left(x-2\right)\left(x+2\right)} by \frac{x-4}{x^{2}-4x+4} by multiplying \frac{-x+4}{\left(x-2\right)\left(x+2\right)} by the reciprocal of \frac{x-4}{x^{2}-4x+4}.

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

Extract the negative sign in -x+4.

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

Cancel out x-4 in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Cancel out x-2 in both numerator and denominator.

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

Expand the expression.

$$\frac{-x+2}{x+2}$$

Expand

$$-\frac{x-2}{x+2}$$

Short Solution Steps

Factor x^{2}-4.

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

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

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

Since \frac{x+8}{\left(x-2\right)\left(x+2\right)} and \frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in x+8-2\left(x+2\right).

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

Combine like terms in x+8-2x-4.

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

Divide \frac{-x+4}{\left(x-2\right)\left(x+2\right)} by \frac{x-4}{x^{2}-4x+4} by multiplying \frac{-x+4}{\left(x-2\right)\left(x+2\right)} by the reciprocal of \frac{x-4}{x^{2}-4x+4}.

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

Extract the negative sign in -x+4.

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

Cancel out x-4 in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Cancel out x-2 in both numerator and denominator.

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

Expand the expression.

$$\frac{-x+2}{x+2}$$