Lời giải:
ĐKXĐ: $x\geq \frac{1}{2}$
PT $\Leftrightarrow (\sqrt{2x-1}-3)+(\sqrt{x+4}-3)=\sqrt{3x+1}-4$
$\Leftrightarrow \frac{2x-1-9}{\sqrt{2x-1}+3}+\frac{x+4-9}{\sqrt{x+4}+3}=\frac{3x+1-16}{\sqrt{3x+1}+4}$
$\Leftrightarrow \frac{2(x-5)}{\sqrt{2x-1}+3}+\frac{x-5}{\sqrt{x+4}+3}=\frac{3(x-5)}{\sqrt{3x+1}+4}$
$\Leftrightarrow (x-5)\left[\frac{2}{\sqrt{2x-1}+3}+\frac{1}{\sqrt{x+4}+3}-\frac{3}{\sqrt{3x+1}+4}\right]=0$
Dễ thấy với mọi $x\geq \frac{1}{2}$ thì:
$0< \sqrt{2x-1}+3< \sqrt{3x+1}+4; 0< \sqrt{x+4}+3< \sqrt{3x+1}+4$
Do đó $\frac{2}{\sqrt{2x-1}+3}>\frac{2}{\sqrt{3x+1}+4}; \frac{1}{\sqrt{x+4}+3}> \frac{1}{\sqrt{3x+1}+4}$
$\Rightarrow \frac{2}{\sqrt{2x-1}+3}+\frac{1}{\sqrt{x+4}+3}-\frac{3}{\sqrt{3x+1}+4}>0$
hay giá trị biểu thức trong ngoặc [ ] khác $0$
$\Rightarrow x-5=0\Rightarrow x=5$ là nghiệm duy nhất
