\(\dfrac{1}{\sqrt{x^2-5x+6}}=\dfrac{1}{\sqrt{x^2-2x-3x+6}}=\dfrac{1}{\sqrt{x\left(x-2\right)-3\left(x-2\right)}}=\dfrac{1}{\sqrt{\left(x-2\right)\left(x-3\right)}}\)
Để bt có nghĩa thì:
\(\sqrt{\left(x-2\right)\left(x-3\right)}>0\)
\(\Rightarrow\left(x-2\right)\left(x-3\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< 3\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< 2\end{matrix}\right.\)
Vậy x>3 hoặc x<2
