Solve (x-2/x+1+5-x/x-2):[(1/x^2-x-2-1/x^2+3x+2)*(frac{x+1}{x}+frac{3-x^2}{x^2+x})]

$$( \frac { x - 2 } { x + 1 } + \frac { 5 - x } { x - 2 } ) : [ ( \frac { 1 } { x ^ { 2 } - x - 2 } - \frac { 1 } { x ^ { 2 } + 3 x + 2 } ) \cdot ( \frac { x + 1 } { x } + \frac { 3 - x ^ { 2 } } { x ^ { 2 } + x } ) ]$$

Evaluate

$$\frac{9x\left(x+1\right)}{8}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+1 and x-2 is \left(x-2\right)\left(x+1\right). Multiply \frac{x-2}{x+1} times \frac{x-2}{x-2}. Multiply \frac{5-x}{x-2} times \frac{x+1}{x+1}.

$$\frac{\frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+1\right)}+\frac{\left(5-x\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

Since \frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+1\right)} and \frac{\left(5-x\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\left(x-2\right)\left(x-2\right)+\left(5-x\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

Do the multiplications in \left(x-2\right)\left(x-2\right)+\left(5-x\right)\left(x+1\right).

$$\frac{\frac{x^{2}-2x-2x+4+5x+5-x^{2}-x}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

Combine like terms in x^{2}-2x-2x+4+5x+5-x^{2}-x.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

Factor x^{2}-x-2. Factor x^{2}+3x+2.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{\left(x-2\right)\left(x+1\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

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

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{x+2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

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

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{x+2-\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

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

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{x+2-x+2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

Combine like terms in x+2-x+2.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

Factor x^{2}+x.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x\left(x+1\right)}\right)}$$

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

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{\left(x+1\right)\left(x+1\right)}{x\left(x+1\right)}+\frac{3-x^{2}}{x\left(x+1\right)}\right)}$$

Since \frac{\left(x+1\right)\left(x+1\right)}{x\left(x+1\right)} and \frac{3-x^{2}}{x\left(x+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\times \frac{\left(x+1\right)\left(x+1\right)+3-x^{2}}{x\left(x+1\right)}}$$

Do the multiplications in \left(x+1\right)\left(x+1\right)+3-x^{2}.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\times \frac{x^{2}+x+1+x+3-x^{2}}{x\left(x+1\right)}}$$

Combine like terms in x^{2}+x+1+x+3-x^{2}.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\times \frac{2x+4}{x\left(x+1\right)}}$$

Multiply \frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)} times \frac{2x+4}{x\left(x+1\right)} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4\left(2x+4\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)}}$$

Divide \frac{9}{\left(x-2\right)\left(x+1\right)} by \frac{4\left(2x+4\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)} by multiplying \frac{9}{\left(x-2\right)\left(x+1\right)} by the reciprocal of \frac{4\left(2x+4\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)}.

$$\frac{9\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)}{\left(x-2\right)\left(x+1\right)\times 4\left(2x+4\right)}$$

Cancel out \left(x-2\right)\left(x+1\right) in both numerator and denominator.

$$\frac{9x\left(x+1\right)\left(x+2\right)}{4\left(2x+4\right)}$$

Factor the expressions that are not already factored.

$$\frac{9x\left(x+1\right)\left(x+2\right)}{2\times 4\left(x+2\right)}$$

Cancel out x+2 in both numerator and denominator.

$$\frac{9x\left(x+1\right)}{2\times 4}$$

Expand the expression.

$$\frac{9x^{2}+9x}{8}$$

Expand

$$\frac{9x^{2}+9x}{8}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+1 and x-2 is \left(x-2\right)\left(x+1\right). Multiply \frac{x-2}{x+1} times \frac{x-2}{x-2}. Multiply \frac{5-x}{x-2} times \frac{x+1}{x+1}.

$$\frac{\frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+1\right)}+\frac{\left(5-x\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

Since \frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+1\right)} and \frac{\left(5-x\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\left(x-2\right)\left(x-2\right)+\left(5-x\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

Do the multiplications in \left(x-2\right)\left(x-2\right)+\left(5-x\right)\left(x+1\right).

$$\frac{\frac{x^{2}-2x-2x+4+5x+5-x^{2}-x}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

Combine like terms in x^{2}-2x-2x+4+5x+5-x^{2}-x.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

Factor x^{2}-x-2. Factor x^{2}+3x+2.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{\left(x-2\right)\left(x+1\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

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

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{x+2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

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

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{x+2-\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

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

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{x+2-x+2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

Combine like terms in x+2-x+2.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}$$

Factor x^{2}+x.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x\left(x+1\right)}\right)}$$

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

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{\left(x+1\right)\left(x+1\right)}{x\left(x+1\right)}+\frac{3-x^{2}}{x\left(x+1\right)}\right)}$$

Since \frac{\left(x+1\right)\left(x+1\right)}{x\left(x+1\right)} and \frac{3-x^{2}}{x\left(x+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\times \frac{\left(x+1\right)\left(x+1\right)+3-x^{2}}{x\left(x+1\right)}}$$

Do the multiplications in \left(x+1\right)\left(x+1\right)+3-x^{2}.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\times \frac{x^{2}+x+1+x+3-x^{2}}{x\left(x+1\right)}}$$

Combine like terms in x^{2}+x+1+x+3-x^{2}.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\times \frac{2x+4}{x\left(x+1\right)}}$$

Multiply \frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)} times \frac{2x+4}{x\left(x+1\right)} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4\left(2x+4\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)}}$$

Divide \frac{9}{\left(x-2\right)\left(x+1\right)} by \frac{4\left(2x+4\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)} by multiplying \frac{9}{\left(x-2\right)\left(x+1\right)} by the reciprocal of \frac{4\left(2x+4\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)}.

$$\frac{9\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)}{\left(x-2\right)\left(x+1\right)\times 4\left(2x+4\right)}$$

Cancel out \left(x-2\right)\left(x+1\right) in both numerator and denominator.

$$\frac{9x\left(x+1\right)\left(x+2\right)}{4\left(2x+4\right)}$$

Factor the expressions that are not already factored.

$$\frac{9x\left(x+1\right)\left(x+2\right)}{2\times 4\left(x+2\right)}$$

Cancel out x+2 in both numerator and denominator.

$$\frac{9x\left(x+1\right)}{2\times 4}$$

Expand the expression.

$$\frac{9x^{2}+9x}{8}$$