\(\dfrac{x+1}{x+2+m}=\dfrac{x-1}{x+2-m}\\ ĐKXĐ:\left\{{}\begin{matrix}x\ne m-2\\x\ne-m-2\end{matrix}\right.\\ \Rightarrow\dfrac{\left(x+1\right)\left(x+2-m\right)}{\left(x+2+m\right)\left(x+2-m\right)}=\dfrac{\left(x-1\right)\left(x+2+m\right)}{\left(x+2+m\right)\left(x+2-m\right)}\)
\(\Rightarrow x^2+2x-mx+x+2-m=x^2+2x+mx-x-2-m\\ \Leftrightarrow x^2+2x-mx+x-m-x^2-2x-mx+x+m=-2-2\\ \Leftrightarrow-2mx+2x=-4\\ \Leftrightarrow-2x\left(m-1\right)=-4\)
+) Với \(m\ne1\Leftrightarrow x=\dfrac{2}{m-1}\)
Khi đó : \(\left\{{}\begin{matrix}\dfrac{2}{m-1}\ne m-2\\\dfrac{2}{m-1}\ne-m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{m-1}\ne\dfrac{\left(m-2\right)\left(m-1\right)}{m-1}\\\dfrac{2}{m-1}\ne\dfrac{\left(-m-2\right)\left(m-1\right)}{m-1}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\ne m^2-2m-m+2\\2\ne-m^2-2m+m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2-3m\ne0\\m^2+m\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\left(m-3\right)\ne0\\m\left(m+1\right)\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ne0\\m-3\ne0\\m\ne0\\m+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\ne0\\m\ne3\\m\ne-1\end{matrix}\right.\)
+) Với \(m=1\Leftrightarrow0x=-4\left(Vô\text{ lý }\right)\)
\(\Rightarrow S=\varnothing\)
Vậy với \(m\ne0;m\ne\pm1;m\ne3\), pt có 1 nghiệm là \(x=\dfrac{2}{m-1}\)
Với \(m=1\), pt vô nghiệm