Solve a^2-b^2/a^2+2ab+b^2div2a-2b/a+b

$$\frac { a ^ { 2 } - b ^ { 2 } } { a ^ { 2 } + 2 a b + b ^ { 2 } } \div \frac { 2 a - 2 b } { a + b }$$

Evaluate

$$\frac{1}{2}=0.5$$

Short Solution Steps

Divide \frac{a^{2}-b^{2}}{a^{2}+2ab+b^{2}} by \frac{2a-2b}{a+b} by multiplying \frac{a^{2}-b^{2}}{a^{2}+2ab+b^{2}} by the reciprocal of \frac{2a-2b}{a+b}.

$$\frac{\left(a^{2}-b^{2}\right)\left(a+b\right)}{\left(a^{2}+2ab+b^{2}\right)\left(2a-2b\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(a-b\right)\left(a+b\right)^{2}}{2\left(a-b\right)\left(a+b\right)^{2}}$$

Cancel out \left(a-b\right)\left(a+b\right)^{2} in both numerator and denominator.

$$\frac{1}{2}$$

Factor

$$\frac{1}{2} = 0.5$$