Để A>-1 thì A+1>0
\(\Leftrightarrow\dfrac{x^2+1}{x-1}+1>0\)
\(\Leftrightarrow\dfrac{x^2+1+x-1}{x-1}>0\)
\(\Leftrightarrow\dfrac{x^2+x}{x-1}>0\)
\(\Leftrightarrow x\cdot\dfrac{x+1}{x-1}>0\)
Trường hợp 1:
\(\left\{{}\begin{matrix}x>0\\\dfrac{x+1}{x-1}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x>1\)
Trường hợp 2:
\(\left\{{}\begin{matrix}x< 0\\\dfrac{x+1}{x-1}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0\\\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< 0\\-1< x< 1\end{matrix}\right.\Leftrightarrow-1< x< 0\)
Vậy: Khi x>1 hoặc -1<x<0 thì A>-1
