\(\dfrac{\left|x+1\right|}{x+1}=1\)
\(\Leftrightarrow\left|x+1\right|=x+1\)( ĐK: \(x\ge-1\))
\(\Leftrightarrow\left[{}\begin{matrix}x+1=x+1\\x+1=-x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1-x-1=0\\x+1+x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0x=0\\2x+2=0\end{matrix}\right.\)(luôn đúng)
\(\Leftrightarrow\left[{}\begin{matrix}0x=0\\x=-1\left(l\right)\end{matrix}\right.\)
=> pt vô số nghiệm ( loại x =- 1)