Lời giải:

ĐK \(x\geq \frac{1}{2}\)

Ta có:

\(2\sqrt{2x-1}+\sqrt{x+3}-\sqrt{5x+11}=0\)

\(\Leftrightarrow 2(\sqrt{2x-1}-1)+(\sqrt{x+3}-2)-(\sqrt{5x+11}-4)=0\)

\(\Leftrightarrow 2. \frac{2x-2}{\sqrt{2x-1}+1}+\frac{x-1}{\sqrt{x+3}+2}-\frac{5x-5}{\sqrt{5x+11}+4}=0\)

\(\Leftrightarrow (x-1)\left[ \frac{4}{\sqrt{2x-1}+1}+\frac{1}{\sqrt{x+3}+2}-\frac{5}{\sqrt{5x+11}+4}\right]=0\) (*)

Ta thấy với \(x\geq \frac{1}{2} \Rightarrow \left\{\begin{matrix} 2x-1< 5x+11\\ x+3< 5x+11\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} \sqrt{2x-1}+1< \sqrt{5x+11}+4\\ \sqrt{x+3}+2< \sqrt{5x+11}+4\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} \frac{4}{\sqrt{2x-1}+1}>\frac{4}{\sqrt{5x+11}+4}\\ \frac{1}{\sqrt{x+3}+2}> \frac{1}{\sqrt{5x+11}+4}\end{matrix}\right.\)

Do đó biểu thức trong ngoặc vuông luôn lớn hơn 0

Suy ra từ (*) ta có: \(x-1=0\Leftrightarrow x=1\)