Khi giải phương trình \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{x^2+4}{x^2-2x-3}\) , ta được nghiệm là
\(x=-4\).\(x=4\).\(x=2\).\(x=-2\).Hướng dẫn giải:\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{x^2+4}{x^2-2x-3}\) (ĐK: \(\left\{{}\begin{matrix}x\ne3\\x\ne-1\end{matrix}\right.\) )
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{x^2+4}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow\)\(x\left(x+1\right)+x\left(x-3\right)=2x^2+8\)
\(\Leftrightarrow-2x=8\)
\(\Leftrightarrow x=-4\) (tmđk)