ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow3\sqrt{x}+\sqrt{7\left(x-2\right)}=x+7\)
Ta có:
\(3\sqrt{x}+\sqrt{7\left(x-2\right)}\le\dfrac{1}{2}\left(9+x\right)+\dfrac{1}{2}\left(7+x-2\right)=x+7\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\sqrt{x}=3\\7=x-2\end{matrix}\right.\) \(\Leftrightarrow x=9\)

