1: ĐKXĐ: \(x\notin\left\{0;-1;3\right\}\)
\(A=\dfrac{2x^2+x^2-1-3x^2+2x+1}{x\left(x+1\right)}:\dfrac{-x+5+x-3}{x-3}\)
\(=\dfrac{2x}{x\left(x+1\right)}\cdot\dfrac{x-3}{2}\)
\(=\dfrac{x-3}{x+1}\)
2: Ta có: |x+2|=1
=>\(\left[{}\begin{matrix}x+2=1\\x+2=-1\end{matrix}\right.\Leftrightarrow x=-3\)
Thay x=-3 vào A, ta được:
\(A=\dfrac{-3-3}{-3+1}=\dfrac{-6}{-2}=3\)


