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

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

Evaluate

$$2$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{x+2}{x+2}.

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

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

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

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

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

Combine like terms in 2x+4-8.

$$\frac{\frac{2x-4}{x+2}}{\frac{x^{2}-4x+4}{x^{2}-4}}$$

Factor the expressions that are not already factored in \frac{x^{2}-4x+4}{x^{2}-4}.

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

Cancel out x-2 in both numerator and denominator.

$$\frac{\frac{2x-4}{x+2}}{\frac{x-2}{x+2}}$$

Divide \frac{2x-4}{x+2} by \frac{x-2}{x+2} by multiplying \frac{2x-4}{x+2} by the reciprocal of \frac{x-2}{x+2}.

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

Cancel out x+2 in both numerator and denominator.

$$\frac{2x-4}{x-2}$$

Factor the expressions that are not already factored.

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

Cancel out x-2 in both numerator and denominator.

$$2$$

Factor

$$2$$