\(\left(x+\frac{1}{2}\right)^4=\frac{16}{81}\)
\(\Rightarrow\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\\\left(x+\frac{1}{2}\right)^4=\left(\frac{-2}{3}\right)^4\end{cases}}\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{2}{3}\\x+\frac{1}{2}=\frac{-2}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=\frac{-7}{6}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{6};\frac{-7}{6}\right\}\)
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