Question

Rút gọn biểu thức và tìm điều kiện xác định

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

Answer 1

ĐKXĐ: x ≠ 2; x ≠ -2 ;

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

\(=\left(\dfrac{1}{x+2}+\dfrac{x}{\left(x+2\right)\left(x-2\right)}+\dfrac{-2}{x-2}\right):\left(1-\dfrac{x}{x+2}\right)\)

\(=\left(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}+\dfrac{x}{\left(x+2\right)\left(x-2\right)}+\dfrac{-2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\right):\left(1-\dfrac{x}{x+2}\right)\)
\(=\dfrac{-6}{\left(x+2\right)\left(x-2\right)}:\left(\dfrac{x+2}{x+2}-\dfrac{x}{x+2}\right)\)

= \(\dfrac{-6}{\left(x+2\right)\left(x-2\right)}:\dfrac{2}{x+2}\)

= \(\dfrac{-6}{\left(x+2\right)\left(x-2\right)}.\dfrac{x+2}{2}\)

\(=\dfrac{-3}{x-2}=\dfrac{3}{2-x}\)