\(\Leftrightarrow\dfrac{x+3}{x-3}-\dfrac{1}{x}-\dfrac{3}{x\left(x-3\right)}=0\\ ĐKXĐ:\left\{{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\\ \Leftrightarrow x\left(x+3\right)-\left(x-3\right)-3=0\)
\(\Leftrightarrow x^2+3x-x+3-3=0\\\Leftrightarrow x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=-2\)
Ta có: \(\dfrac{x+3}{x-3}-\dfrac{1}{x}-\dfrac{3}{x^2-3x}=0\)
\(\Leftrightarrow x^2+3x-x+3-3=0\)
\(\Leftrightarrow x^2+2x=0\)
\(\Leftrightarrow x=-2\)
