\(\dfrac{3x-5}{\sqrt{x+4}}=\sqrt{x+4}\) (ĐK: \(x>-4\) )
\(\Leftrightarrow3x-5=\sqrt{x+4}\cdot\sqrt{x+4}\)
\(\Leftrightarrow3x-5=\left(\sqrt{x+4}\right)^2\)
\(\Leftrightarrow3x-5=x+4\)
\(\Leftrightarrow3x-x=4+5\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)
Vậy: \(x=\dfrac{9}{2}\)