\(\left(x-3\right)\sqrt{x^2-4}=x^2-9\) (ĐK: \(\left\{{}\begin{matrix}x< -2\\x\ge2\end{matrix}\right.\))
\(\Leftrightarrow\left(x-3\right)\sqrt{x^2-4}=\left(x-3\right)\left(x+3\right)\)
\(\Leftrightarrow\sqrt{x^2-4}=\dfrac{\left(x-3\right)\left(x+3\right)}{x-3}\)
\(\Leftrightarrow\sqrt{x^2-4}=x+3\)
\(\Leftrightarrow x^2-4=\left(x+3\right)^2\)
\(\Leftrightarrow x^2-4=x^2+6x+9\)
\(\Leftrightarrow x^2-x^2-6x=9+4\)
\(\Leftrightarrow-6x=13\)
\(\Leftrightarrow x=-\dfrac{13}{6}\left(tm\right)\)
Vậy: ...