\(\left(x+\frac{1}{6}\right)^2-81=0\\ \Rightarrow\left(x+\frac{1}{6}\right)^2=81\\ \Rightarrow\left(x+\frac{1}{6}\right)^2=9^2=\left(-9\right)^2\\ \Rightarrow\left[{}\begin{matrix}x+\frac{1}{6}=9\\x+\frac{1}{6}=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9-\frac{1}{6}\\x=-9-\frac{1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{54}{6}-\frac{1}{6}\\x=\frac{-54}{6}-\frac{1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{53}{6}\\x=\frac{-55}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{53}{6};\frac{-55}{6}\right\}\)
\(\left(x+\frac{1}{6}\right)^2-81=0\)
\(\Rightarrow\left(x+\frac{1}{6}\right)^2=0+81\)
\(\Rightarrow\left(x+\frac{1}{6}\right)^2=81\)
\(\Rightarrow x+\frac{1}{6}=\pm9.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{6}=9\\x+\frac{1}{6}=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9-\frac{1}{6}\\x=\left(-9\right)-\frac{1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{53}{6}\\x=-\frac{55}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{53}{6};-\frac{55}{6}\right\}.\)
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