ĐKXĐ: ...
\(\Leftrightarrow\sqrt{x+3}+\sqrt{5-x}=2x-4\sqrt{x}+6\)
\(\Leftrightarrow\sqrt{x+3}+\sqrt{5-x}=2\left(\sqrt{x}-1\right)^2+4\)
Ta có: \(VT\le\sqrt{\left(1+1\right)\left(x+3+5-x\right)}=4\)
\(VP\ge4\Rightarrow VT\le VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\sqrt{x+3}=\sqrt{5-x}\\\sqrt{x}-1=0\end{matrix}\right.\)
\(\Rightarrow x=1\)