Lời giải:
ĐK: \(x\neq \left\{0; \pm 2\right\}\)
Ta có:
\(B=\left ( \frac{1}{x^2+2x}-\frac{2}{x^2-2x}+\frac{1}{x^2-4} \right ):\left(\frac{1}{x-2}-\frac{1}{x}\right)\)
\(B=\left ( \frac{x-2}{(x^2+2x)(x-2)}-\frac{2(x+2)}{(x^2-2x)(x+2)}+\frac{x}{x(x^2-4)} \right ):\left ( \frac{x-(x-2)}{x(x-2)} \right )\)
\(B=\left ( \frac{x-2}{x(x^2-4)}-\frac{2x+4}{x(x^2-4)}+\frac{x}{x(x^2-4)} \right ):\frac{2}{x(x-2)}\)
\(B=\frac{x-2-2x-4+x}{x(x^2-4)}.\frac{x(x-2)}{2}\)
\(B=\frac{-6}{x(x-2)(x+2)}.\frac{x(x-2)}{2}=\frac{-3}{x+2}\)
\(B=\left(\dfrac{1}{x^2+2x}-\dfrac{2}{x^2-2x}+\dfrac{1}{x^2-4}\right):\left(\dfrac{1}{x-2}-\dfrac{1}{x}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{x\left(x+2\right)}-\dfrac{2}{x\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{1}{x-2}-\dfrac{1}{x}\right)\)
\(\Leftrightarrow B=\left(\dfrac{x-2}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\dfrac{x}{x\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}\right)\)
\(\Leftrightarrow B=\dfrac{\left(x-2\right)-2\left(x+2\right)+x}{x\left(x-2\right)\left(x+2\right)}:\dfrac{x-\left(x-2\right)}{x\left(x-2\right)}\)
\(\Leftrightarrow B=\dfrac{x-2-2x-4+x}{x\left(x-2\right)\left(x+2\right)}:\dfrac{x-x+2}{x\left(x-2\right)}\)
\(\Leftrightarrow B=\dfrac{-6}{x\left(x-2\right)\left(x+2\right)}:\dfrac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow B=\dfrac{-6}{x\left(x-2\right)\left(x+2\right)}.\dfrac{x\left(x-2\right)}{2}\)
\(\Leftrightarrow B=\dfrac{-6.x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right).2}\)
\(\Leftrightarrow B=\dfrac{-3}{x+2}\)
