\(\sqrt{x-2}\) = \(\dfrac{2}{3}\) (đk \(x-2\) ≥ 0; \(x\ge\) 2)
\(x-2\) =(\(\dfrac{2}{3}\))2
\(x-2\) = \(\dfrac{4}{9}\)
\(x=\dfrac{4}{9}\) + 2
\(x=\dfrac{22}{9}\)
Vậy \(x=\dfrac{22}{9}\)
ĐKXĐ: \(x\ge2\)
\(\sqrt{x-2}=\dfrac{2}{3}\)
\(x-2=\dfrac{4}{9}\)
\(x=\dfrac{4}{9}+2\)
\(x=\dfrac{22}{9}\) (nhận)
Vậy \(x=\dfrac{22}{9}\)