\(\dfrac{a+1}{x-1}=1-a\) \(\left(đk:a,x>0;a,x< 1\right)\)
\(\Leftrightarrow\dfrac{a+1}{x-1}+1+a=0\)
\(\Leftrightarrow\dfrac{a+1+\left(1+a\right)\left(x-1\right)}{x-1}=0\)
\(\Leftrightarrow a+1+x-1+ax-a=0\)
\(\Leftrightarrow x+ax=0\)
\(\Leftrightarrow x\left(1+a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1+a=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\a=-1\left(ktm\right)\end{matrix}\right.\)
Vậy \(a=0\) thì pt \(\dfrac{a+1}{x-1}=1-a\) nhận giá trị nguyên