\(1,ĐKXĐ:x^2-4\ne0\Leftrightarrow x\ne\pm2\\ 2,A=\dfrac{x^2}{x^2-4}-\dfrac{x}{x-2}+\dfrac{2}{x+2}=\dfrac{x^2-x\left(x+2\right)+2\left(x-2\right)}{x^2-4}\\ =\dfrac{x^2-x^2-2x+2x-4}{x^2-4}=\dfrac{-4}{x^2-4}\\ 3,Thay.x=1.vào.A.rút.gọn:A=\dfrac{-4}{x^2-4}=\dfrac{-4}{1^2-4}=\dfrac{-4}{-3}=\dfrac{4}{3}\\ Vậy:A=\dfrac{4}{3}.tại.x=1\)
1: ĐKXĐ: x<>2; x<>-2
2: \(A=\dfrac{x^2-x\left(x+2\right)+2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-x^2-2x+2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{4}{x^2-4}\)
3: Khi x=1 thì \(A=\dfrac{4}{1^2-4}=-\dfrac{4}{3}\)
