Solve x/x+2*x^2+x-2/x^2-8x-x^2-2x+1/x-1divx-8/2=

$$\frac { x } { x + 2 } \times \frac { x ^ { 2 } + x - 2 } { x ^ { 2 } - 8 x } - \frac { x ^ { 2 } - 2 x + 1 } { x - 1 } \div \frac { x - 8 } { 2 } =$$

Evaluate

$$-\frac{x-1}{x-8}$$

Short Solution Steps

Multiply \frac{x}{x+2} times \frac{x^{2}+x-2}{x^{2}-8x} by multiplying numerator times numerator and denominator times denominator.

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

Divide \frac{x^{2}-2x+1}{x-1} by \frac{x-8}{2} by multiplying \frac{x^{2}-2x+1}{x-1} by the reciprocal of \frac{x-8}{2}.

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

Factor the expressions that are not already factored in \frac{\left(x^{2}-2x+1\right)\times 2}{\left(x-1\right)\left(x-8\right)}.

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

Cancel out x-1 in both numerator and denominator.

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

Factor \left(x+2\right)\left(x^{2}-8x\right).

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x\left(x-8\right)\left(x+2\right) and x-8 is x\left(x-8\right)\left(x+2\right). Multiply \frac{2\left(x-1\right)}{x-8} times \frac{x\left(x+2\right)}{x\left(x+2\right)}.

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

Since \frac{x\left(x^{2}+x-2\right)}{x\left(x-8\right)\left(x+2\right)} and \frac{2\left(x-1\right)x\left(x+2\right)}{x\left(x-8\right)\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

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

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

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

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

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

Factor the expressions that are not already factored in \frac{-x^{3}-x^{2}+2x}{x\left(x-8\right)\left(x+2\right)}.

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

Cancel out x\left(x+2\right) in both numerator and denominator.

$$\frac{-x+1}{x-8}$$

Expand

$$-\frac{x-1}{x-8}$$

Short Solution Steps

Multiply \frac{x}{x+2} times \frac{x^{2}+x-2}{x^{2}-8x} by multiplying numerator times numerator and denominator times denominator.

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

Divide \frac{x^{2}-2x+1}{x-1} by \frac{x-8}{2} by multiplying \frac{x^{2}-2x+1}{x-1} by the reciprocal of \frac{x-8}{2}.

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

Factor the expressions that are not already factored in \frac{\left(x^{2}-2x+1\right)\times 2}{\left(x-1\right)\left(x-8\right)}.

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

Cancel out x-1 in both numerator and denominator.

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

Factor \left(x+2\right)\left(x^{2}-8x\right).

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x\left(x-8\right)\left(x+2\right) and x-8 is x\left(x-8\right)\left(x+2\right). Multiply \frac{2\left(x-1\right)}{x-8} times \frac{x\left(x+2\right)}{x\left(x+2\right)}.

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

Since \frac{x\left(x^{2}+x-2\right)}{x\left(x-8\right)\left(x+2\right)} and \frac{2\left(x-1\right)x\left(x+2\right)}{x\left(x-8\right)\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

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

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

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

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

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

Factor the expressions that are not already factored in \frac{-x^{3}-x^{2}+2x}{x\left(x-8\right)\left(x+2\right)}.

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

Cancel out x\left(x+2\right) in both numerator and denominator.

$$\frac{-x+1}{x-8}$$