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

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

Evaluate

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

Short Solution Steps

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

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

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

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

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

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

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

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

Factor x^{2}-1.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+1\right)\left(-x+1\right) and \left(x-1\right)\left(x+1\right) is \left(x-1\right)\left(x+1\right). Multiply \frac{4x}{\left(x+1\right)\left(-x+1\right)} times \frac{-1}{-1}.

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

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

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

Factor the expressions that are not already factored in \frac{-4x-4x^{2}}{\left(x-1\right)\left(x+1\right)}.

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

Extract the negative sign in -1-x.

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

Cancel out x+1 in both numerator and denominator.

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

Factor the expressions that are not already factored in \frac{4x^{2}-4}{x^{2}-2x+1}.

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

Cancel out x-1 in both numerator and denominator.

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

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

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

Cancel out 4\left(x-1\right) in both numerator and denominator.

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

Expand

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

Short Solution Steps

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

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

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

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

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

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

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

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

Factor x^{2}-1.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+1\right)\left(-x+1\right) and \left(x-1\right)\left(x+1\right) is \left(x-1\right)\left(x+1\right). Multiply \frac{4x}{\left(x+1\right)\left(-x+1\right)} times \frac{-1}{-1}.

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

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

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

Factor the expressions that are not already factored in \frac{-4x-4x^{2}}{\left(x-1\right)\left(x+1\right)}.

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

Extract the negative sign in -1-x.

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

Cancel out x+1 in both numerator and denominator.

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

Factor the expressions that are not already factored in \frac{4x^{2}-4}{x^{2}-2x+1}.

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

Cancel out x-1 in both numerator and denominator.

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

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

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

Cancel out 4\left(x-1\right) in both numerator and denominator.

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