a) Ta có: \(\left(x-1\right)^2+2=x^2+3x\)
\(\Leftrightarrow x^2-2x+1+2-x^2-3x=0\)
\(\Leftrightarrow-5x=-3\)
hay \(x=\dfrac{3}{5}\)
ĐKXĐ:\(x\ne\pm3\)
\(\dfrac{x-3}{x+3}-\dfrac{2}{x-3}=\dfrac{-3\left(x-1\right)}{x^2-9}\)
⇔\(\left(x-3\right)\left(x-3\right)-2\left(x+3\right)+3\left(x-1\right)=0\)
⇔\(x^2-3x-3x+9-2x-6+3x-3=0\)
⇔\(x^2-3x=0\)
⇔\(x\left(x-3\right)=0\)
⇔\(\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x=0\\x=3\left(\notinĐKXĐ\right)\end{matrix}\right.\)