Solve (1-2/x)^2divx^2-4x+4/x^2-4-x+4/x+2

$$( 1 - \frac { 2 } { x } ) ^ { 2 } \div \frac { x ^ { 2 } - 4 x + 4 } { x ^ { 2 } - 4 } - \frac { x + 4 } { x + 2 }$$

Evaluate

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

Short Solution Steps

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

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

Since \frac{x}{x} and \frac{2}{x} have the same denominator, subtract them by subtracting their numerators.

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

To raise \frac{x-2}{x} to a power, raise both numerator and denominator to the power and then divide.

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

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

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

Cancel out x-2 in both numerator and denominator.

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

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

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

Cancel out x-2 in both numerator and denominator.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x^{2} and x+2 is \left(x+2\right)x^{2}. Multiply \frac{\left(x-2\right)\left(x+2\right)}{x^{2}} times \frac{x+2}{x+2}. Multiply \frac{x+4}{x+2} times \frac{x^{2}}{x^{2}}.

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

Since \frac{\left(x-2\right)\left(x+2\right)\left(x+2\right)}{\left(x+2\right)x^{2}} and \frac{\left(x+4\right)x^{2}}{\left(x+2\right)x^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(x-2\right)\left(x+2\right)\left(x+2\right)-\left(x+4\right)x^{2}.

$$\frac{x^{3}+4x^{2}+4x-2x^{2}-8x-8-x^{3}-4x^{2}}{\left(x+2\right)x^{2}}$$

Combine like terms in x^{3}+4x^{2}+4x-2x^{2}-8x-8-x^{3}-4x^{2}.

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

Expand \left(x+2\right)x^{2}.

$$\frac{-2x^{2}-4x-8}{x^{3}+2x^{2}}$$

Expand

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

Short Solution Steps

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

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

Since \frac{x}{x} and \frac{2}{x} have the same denominator, subtract them by subtracting their numerators.

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

To raise \frac{x-2}{x} to a power, raise both numerator and denominator to the power and then divide.

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

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

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

Cancel out x-2 in both numerator and denominator.

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

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

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

Cancel out x-2 in both numerator and denominator.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x^{2} and x+2 is \left(x+2\right)x^{2}. Multiply \frac{\left(x-2\right)\left(x+2\right)}{x^{2}} times \frac{x+2}{x+2}. Multiply \frac{x+4}{x+2} times \frac{x^{2}}{x^{2}}.

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

Since \frac{\left(x-2\right)\left(x+2\right)\left(x+2\right)}{\left(x+2\right)x^{2}} and \frac{\left(x+4\right)x^{2}}{\left(x+2\right)x^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(x-2\right)\left(x+2\right)\left(x+2\right)-\left(x+4\right)x^{2}.

$$\frac{x^{3}+4x^{2}+4x-2x^{2}-8x-8-x^{3}-4x^{2}}{\left(x+2\right)x^{2}}$$

Combine like terms in x^{3}+4x^{2}+4x-2x^{2}-8x-8-x^{3}-4x^{2}.

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

Expand \left(x+2\right)x^{2}.

$$\frac{-2x^{2}-4x-8}{x^{3}+2x^{2}}$$