Solve 1/x^2-4x+4-x/x^2-4+1/2x+4

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

Evaluate

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

Short Solution Steps

Factor x^{2}-4x+4. Factor x^{2}-4.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-2\right)^{2} and \left(x-2\right)\left(x+2\right) is \left(x+2\right)\left(x-2\right)^{2}. Multiply \frac{1}{\left(x-2\right)^{2}} times \frac{x+2}{x+2}. Multiply \frac{x}{\left(x-2\right)\left(x+2\right)} times \frac{x-2}{x-2}.

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

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

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

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

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

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

$$\frac{3x+2-x^{2}}{\left(x+2\right)\left(x-2\right)^{2}}+\frac{1}{2x+4}$$

Factor 2x+4.

$$\frac{3x+2-x^{2}}{\left(x+2\right)\left(x-2\right)^{2}}+\frac{1}{2\left(x+2\right)}$$

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)^{2} and 2\left(x+2\right) is 2\left(x+2\right)\left(x-2\right)^{2}. Multiply \frac{3x+2-x^{2}}{\left(x+2\right)\left(x-2\right)^{2}} times \frac{2}{2}. Multiply \frac{1}{2\left(x+2\right)} times \frac{\left(x-2\right)^{2}}{\left(x-2\right)^{2}}.

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

Since \frac{2\left(3x+2-x^{2}\right)}{2\left(x+2\right)\left(x-2\right)^{2}} and \frac{\left(x-2\right)^{2}}{2\left(x+2\right)\left(x-2\right)^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 2\left(3x+2-x^{2}\right)+\left(x-2\right)^{2}.

$$\frac{6x+4-2x^{2}+x^{2}-4x+4}{2\left(x+2\right)\left(x-2\right)^{2}}$$

Combine like terms in 6x+4-2x^{2}+x^{2}-4x+4.

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

Factor the expressions that are not already factored in \frac{2x+8-x^{2}}{2\left(x+2\right)\left(x-2\right)^{2}}.

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

Extract the negative sign in -2-x.

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

Cancel out x+2 in both numerator and denominator.

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

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

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

To find the opposite of x-4, find the opposite of each term.

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

Factor

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