Solve (-3x^2/x^2-1+4)*(x+1/x^2-4)=

$$( \frac { - 3 x ^ { 2 } } { x ^ { 2 } - 1 } + 4 ) \cdot ( \frac { x + 1 } { x ^ { 2 } - 4 } ) =$$

Evaluate

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

Short Solution Steps

Factor x^{2}-1.

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

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

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

Since \frac{-3x^{2}}{\left(x-1\right)\left(x+1\right)} and \frac{4\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in -3x^{2}+4\left(x-1\right)\left(x+1\right).

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

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

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

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

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

Cancel out \left(x+1\right)\left(x^{2}-4\right) in both numerator and denominator.

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

Expand

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

Short Solution Steps

Factor x^{2}-1.

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

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

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

Since \frac{-3x^{2}}{\left(x-1\right)\left(x+1\right)} and \frac{4\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in -3x^{2}+4\left(x-1\right)\left(x+1\right).

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

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

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

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

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

Cancel out \left(x+1\right)\left(x^{2}-4\right) in both numerator and denominator.

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