ĐK: \(x\ge\frac{1}{3}\)
Pt đã cho tương đương với \(\left(18x^2-2x-\frac{8}{3}\right)+9\left(\sqrt{x-\frac{1}{3}}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left(18x-8\right)\left(x+\frac{1}{3}\right)+9\frac{x-\frac{1}{3}-\frac{1}{9}}{\sqrt{x-\frac{1}{3}}+\frac{1}{3}}=0\)
\(\Leftrightarrow\left(x-\frac{4}{9}\right)\text{[}18\left(x+\frac{1}{3}\right)+9\frac{1}{\sqrt{x-\frac{1}{3}}+\frac{1}{2}}\text{]}=0\Rightarrow x=\frac{4}{9}\)
CM: Với \(x\ge\frac{1}{3}\Rightarrow18\left(x+\frac{1}{3}\right)+9\frac{1}{\sqrt{x-\frac{1}{3}}+\frac{1}{3}}>0\)
Pt đã cho có nghiệm \(x=\frac{4}{9}\)
