Solve F(x)=x^2-x/x-1-(x+1/x-1-x-1/x+1):4/x^2

$$F ( x ) = \frac { x ^ { 2 } - x } { x - 1 } - ( \frac { x + 1 } { x - 1 } - \frac { x - 1 } { x + 1 } ) : \frac { 4 } { x ^ { 2 } }$$

Evaluate

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{x^{2}-x}{x-1}.

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

Cancel out x-1 in both numerator and denominator.

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

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

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

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

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

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

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

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

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

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

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

Cancel out 4 in both numerator and denominator.

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

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

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

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

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

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

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

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

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

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

Expand \left(x-1\right)\left(x+1\right).

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

Expand

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{x^{2}-x}{x-1}.

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

Cancel out x-1 in both numerator and denominator.

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

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

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

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

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

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

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

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

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

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

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

Cancel out 4 in both numerator and denominator.

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

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

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

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

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

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

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

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

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

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

Expand \left(x-1\right)\left(x+1\right).

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