Question

\(B=\left(\dfrac{1}{2-x}+\dfrac{3x}{x^2-4}-\dfrac{2}{2+x}\right):\left(\dfrac{x^2+4}{4-x^2}+1\right)\)

Answer 1

\(B=\left(\dfrac{1}{2-x}+\dfrac{3x}{x^2-4}-\dfrac{2}{2+x}\right)+\left(\dfrac{x^2+4}{4-x^2}+1\right)\)

\(=\left(\dfrac{1}{-\left(x-2\right)}+\dfrac{3x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{2+x}\right)+\dfrac{x^2+4+4-x^2}{4-x^2}\)

\(=\left(-\dfrac{1}{x-2}+\dfrac{3x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{2+x}\right):\dfrac{8}{4-x^2}\)

\(=\dfrac{-\left(x+2\right)+3x-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{4-x^2}{8}\)

\(=\dfrac{-x-2+3x-2x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(2-x\right)\left(2+x\right)}{8}\)

\(=\dfrac{0+2}{x-2}\cdot\dfrac{-\left(x-2\right)}{8}\)

\(=2\cdot\left(-\dfrac{1}{8}\right)\)

\(=-\dfrac{1}{4}\)