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

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

Evaluate

$$9$$

Short Solution Steps

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

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

Since \frac{x+1}{x+1} and \frac{2-x}{x+1} have the same denominator, add them by adding their numerators.

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

Combine like terms in x+1+2-x.

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

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

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

Cancel out x-1 in both numerator and denominator.

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

Divide \frac{3}{x+1} by \frac{1}{3\left(x+1\right)} by multiplying \frac{3}{x+1} by the reciprocal of \frac{1}{3\left(x+1\right)}.

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

Cancel out x+1 in both numerator and denominator.

$$3\times 3$$

Multiply 3 and 3 to get 9.

$$9$$

Factor

$$3^{2}$$