a) Ta có:x(x-1)=-x(x+3)
\(\Leftrightarrow x^2-x=-x^2-3x\)
\(\Leftrightarrow x^2-x+x^2+3x=0\)
\(\Leftrightarrow2x^2+2x=0\)
\(\Leftrightarrow2x\left(x+1\right)=0\)
mà 2≠0
nên \(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy: S={0;-1}
b) ĐKXĐ: \(\left\{{}\begin{matrix}x+1\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)
Ta có: \(\frac{x}{2x-6}-\frac{x}{2x+2}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{4x}{2\left(x+1\right)\left(x-3\right)}\)
Suy ra: \(x\left(x+1\right)-x\left(x-3\right)=4x\)
\(\Leftrightarrow x^2+x-x^2+3x-4x=0\)
\(\Leftrightarrow0x=0\)
Vậy: S={x|x∉{-1;3}}
