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

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

Evaluate

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Short Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

Divide \frac{-x-4}{x\left(x-2\right)} by \frac{x+4}{x^{2}-2x} by multiplying \frac{-x-4}{x\left(x-2\right)} by the reciprocal of \frac{x+4}{x^{2}-2x}.

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

Extract the negative sign in -x-4.

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

Cancel out x+4 in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

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

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Factor

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