ĐKXĐ: \(x\ge\dfrac{1}{2}\)
Nhân 2 vế với \(\sqrt{x+6}-\sqrt{x+2}\) ta được:
\(\sqrt{2x-1}-3=\sqrt{x+6}-\sqrt{x+2}\Leftrightarrow\sqrt{2x-1}-\sqrt{x+6}-\left(3-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\dfrac{2x-1-x-6}{\sqrt{2x-1}+\sqrt{x+6}}-\dfrac{9-x-2}{3+\sqrt{x+2}}=0\)
\(\Leftrightarrow\dfrac{x-7}{\sqrt{2x-1}+\sqrt{x+6}}+\dfrac{x-7}{3+\sqrt{x+2}}=0\)
\(\Leftrightarrow\left(x-7\right)\left(\dfrac{1}{\sqrt{2x-1}+\sqrt{x+6}}+\dfrac{1}{3+\sqrt{x+2}}\right)=0\)
\(\Rightarrow x=7\) (do \(\dfrac{1}{\sqrt{2x-1}+\sqrt{x+6}}+\dfrac{1}{3+\sqrt{x+2}}>0\) \(\forall x\) )
Vậy pt có nghiệm duy nhất \(x=7\)