Solve x+2/x^2-4x+4-1/2x+4-x-1/2x^2-8

$$\frac { x + 2 } { x ^ { 2 } - 4 x + 4 } - \frac { 1 } { 2 x + 4 } - \frac { x - 1 } { 2 x ^ { 2 } - 8 }$$

Evaluate

$$\frac{15x+2}{2\left(x+2\right)\left(x-2\right)^{2}}$$

Short Solution Steps

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

$$\frac{x+2}{\left(x-2\right)^{2}}-\frac{1}{2\left(x+2\right)}-\frac{x-1}{2x^{2}-8}$$

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

$$\frac{\left(x+2\right)\times 2\left(x+2\right)}{2\left(x+2\right)\left(x-2\right)^{2}}-\frac{\left(x-2\right)^{2}}{2\left(x+2\right)\left(x-2\right)^{2}}-\frac{x-1}{2x^{2}-8}$$

Since \frac{\left(x+2\right)\times 2\left(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, subtract them by subtracting their numerators.

$$\frac{\left(x+2\right)\times 2\left(x+2\right)-\left(x-2\right)^{2}}{2\left(x+2\right)\left(x-2\right)^{2}}-\frac{x-1}{2x^{2}-8}$$

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

$$\frac{2x^{2}+4x+4x+8-x^{2}+4x-4}{2\left(x+2\right)\left(x-2\right)^{2}}-\frac{x-1}{2x^{2}-8}$$

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

$$\frac{x^{2}+12x+4}{2\left(x+2\right)\left(x-2\right)^{2}}-\frac{x-1}{2x^{2}-8}$$

Factor 2x^{2}-8.

$$\frac{x^{2}+12x+4}{2\left(x+2\right)\left(x-2\right)^{2}}-\frac{x-1}{2\left(x-2\right)\left(x+2\right)}$$

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

$$\frac{x^{2}+12x+4}{2\left(x+2\right)\left(x-2\right)^{2}}-\frac{\left(x-1\right)\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)^{2}}$$

Since \frac{x^{2}+12x+4}{2\left(x+2\right)\left(x-2\right)^{2}} and \frac{\left(x-1\right)\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{2}+12x+4-\left(x-1\right)\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)^{2}}$$

Do the multiplications in x^{2}+12x+4-\left(x-1\right)\left(x-2\right).

$$\frac{x^{2}+12x+4-x^{2}+2x+x-2}{2\left(x+2\right)\left(x-2\right)^{2}}$$

Combine like terms in x^{2}+12x+4-x^{2}+2x+x-2.

$$\frac{15x+2}{2\left(x+2\right)\left(x-2\right)^{2}}$$

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

$$\frac{15x+2}{2x^{3}-4x^{2}-8x+16}$$

Expand

$$\frac{15x+2}{2\left(x+2\right)\left(x-2\right)^{2}}$$

Short Solution Steps

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

$$\frac{x+2}{\left(x-2\right)^{2}}-\frac{1}{2\left(x+2\right)}-\frac{x-1}{2x^{2}-8}$$

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

$$\frac{\left(x+2\right)\times 2\left(x+2\right)}{2\left(x+2\right)\left(x-2\right)^{2}}-\frac{\left(x-2\right)^{2}}{2\left(x+2\right)\left(x-2\right)^{2}}-\frac{x-1}{2x^{2}-8}$$

Since \frac{\left(x+2\right)\times 2\left(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, subtract them by subtracting their numerators.

$$\frac{\left(x+2\right)\times 2\left(x+2\right)-\left(x-2\right)^{2}}{2\left(x+2\right)\left(x-2\right)^{2}}-\frac{x-1}{2x^{2}-8}$$

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

$$\frac{2x^{2}+4x+4x+8-x^{2}+4x-4}{2\left(x+2\right)\left(x-2\right)^{2}}-\frac{x-1}{2x^{2}-8}$$

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

$$\frac{x^{2}+12x+4}{2\left(x+2\right)\left(x-2\right)^{2}}-\frac{x-1}{2x^{2}-8}$$

Factor 2x^{2}-8.

$$\frac{x^{2}+12x+4}{2\left(x+2\right)\left(x-2\right)^{2}}-\frac{x-1}{2\left(x-2\right)\left(x+2\right)}$$

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

$$\frac{x^{2}+12x+4}{2\left(x+2\right)\left(x-2\right)^{2}}-\frac{\left(x-1\right)\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)^{2}}$$

Since \frac{x^{2}+12x+4}{2\left(x+2\right)\left(x-2\right)^{2}} and \frac{\left(x-1\right)\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{2}+12x+4-\left(x-1\right)\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)^{2}}$$

Do the multiplications in x^{2}+12x+4-\left(x-1\right)\left(x-2\right).

$$\frac{x^{2}+12x+4-x^{2}+2x+x-2}{2\left(x+2\right)\left(x-2\right)^{2}}$$

Combine like terms in x^{2}+12x+4-x^{2}+2x+x-2.

$$\frac{15x+2}{2\left(x+2\right)\left(x-2\right)^{2}}$$

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

$$\frac{15x+2}{2x^{3}-4x^{2}-8x+16}$$