Solve 3-x/x-2div(5/x-2-x-2)*(x+3)

$$\frac { 3 - x } { x - 2 } \div ( \frac { 5 } { x - 2 } - x - 2 ) \cdot ( x + 3 )$$

Evaluate

$$1$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply -x-2 times \frac{x-2}{x-2}.

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

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

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

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

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

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

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

Divide \frac{3-x}{x-2} by \frac{9-x^{2}}{x-2} by multiplying \frac{3-x}{x-2} by the reciprocal of \frac{9-x^{2}}{x-2}.

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

Cancel out x-2 in both numerator and denominator.

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

Factor the expressions that are not already factored in \frac{-x+3}{-x^{2}+9}.

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

Extract the negative sign in 3-x.

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

Cancel out x-3 in both numerator and denominator.

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

Express \frac{-1}{-x-3}\left(x+3\right) as a single fraction.

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

Extract the negative sign in x+3.

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

Cancel out -x-3 in both numerator and denominator.

$$-\left(-1\right)$$

Multiply -1 and -1 to get 1.

$$1$$

Factor

$$1$$