Solve (x+2)*(x-2)-x*(x-1)/x*left(x-2right)^2*x(x-2)/-(x-4)

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

Evaluate

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

Short Solution Steps

Consider \left(x+2\right)\left(x-2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

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

Use the distributive property to multiply x by x-1.

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

To find the opposite of x^{2}-x, find the opposite of each term.

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

Combine x^{2} and -x^{2} to get 0.

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

To find the opposite of x-4, find the opposite of each term.

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

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

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

Extract the negative sign in -4+x.

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

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

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

Expand

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

Short Solution Steps

Consider \left(x+2\right)\left(x-2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

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

Use the distributive property to multiply x by x-1.

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

To find the opposite of x^{2}-x, find the opposite of each term.

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

Combine x^{2} and -x^{2} to get 0.

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

To find the opposite of x-4, find the opposite of each term.

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

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

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

Extract the negative sign in -4+x.

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

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

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