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

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

Solve for x

$$x\in \left(-\frac{1}{2},2\right)$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

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

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

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

Cancel out 2 in both numerator and denominator.

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

Use the distributive property to multiply 2 by -x-1.

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

Subtract \frac{1}{2-x} from both sides.

$$\frac{-2x-2}{x-2}-\frac{1}{2-x}>0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-2 and 2-x is x-2. Multiply \frac{1}{2-x} times \frac{-1}{-1}.

$$\frac{-2x-2}{x-2}-\frac{-1}{x-2}>0$$

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

$$\frac{-2x-2-\left(-1\right)}{x-2}>0$$

Do the multiplications in -2x-2-\left(-1\right).

$$\frac{-2x-2+1}{x-2}>0$$

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

$$\frac{-2x-1}{x-2}>0$$

For the quotient to be positive, -2x-1 and x-2 have to be both negative or both positive. Consider the case when -2x-1 and x-2 are both negative.

$$-2x-1<0 x-2<0$$

The solution satisfying both inequalities is x\in \left(-\frac{1}{2},2\right).

$$x\in \left(-\frac{1}{2},2\right)$$

Consider the case when -2x-1 and x-2 are both positive.

$$x-2>0 -2x-1>0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\frac{1}{2},2\right)$$