Question

\(M=\left(\dfrac{1}{x+2}+\dfrac{2}{2-x}+\dfrac{x}{x^2-4}\right):\left(\dfrac{10-x^2}{x+2}+x-2\right)\)

a) Tìm ĐKXĐ và Rút gọn M

b) Tìm \(x\in Z\) để \(M\in Z\)

Answer 1

TXĐ : \(x\ne\pm2\)

\(M=\left[\dfrac{1}{x+2}-\dfrac{2}{x-2}+\dfrac{x}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{10-x^2+\left(x-2\right)\left(x+2\right)}{x+2}\)

\(=\dfrac{x-2-2\left(x+2\right)+x}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{10-x^2+x^2-1}\)

\(=\dfrac{x-2-2x-4+x}{x-2}.\dfrac{1}{6}\)

\(=\dfrac{-6}{x-2}.\dfrac{1}{6}=\dfrac{1}{2-x}\)