\(\left(\frac{1}{3}x+\frac{1}{2}\right)^2=\frac{4}{81}\)
\(\left(\frac{1}{3}x+\frac{1}{2}\right)^2=\left(\frac{2}{9}\right)^2=\left(\frac{-2}{9}\right)^2\)
=> \(\orbr{\begin{cases}\frac{1}{3}x+\frac{1}{2}=\frac{2}{9}\\\frac{1}{3}x+\frac{1}{2}=\frac{-2}{9}\end{cases}}\)=> \(\orbr{\begin{cases}\frac{1}{3}x=\frac{-5}{18}\\\frac{1}{3}x=\frac{-13}{18}\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{-5}{6}\\x=\frac{-13}{6}\end{cases}}\)
Vậy \(x\in\left\{\frac{-5}{6};\frac{-13}{6}\right\}\)
(1/3x + 1/2)2 = (2/9)2
1/3x+1/2 = 2/9
1/3x = -5/18
x= -5/6