`#3107.101107`
\(\left(x+\dfrac{1}{3}\right)^2=\left(\pm\dfrac{2}{3}\right)^6\\ \left(x+\dfrac{1}{3}\right)^2=\left[\pm\left(\dfrac{2}{3}\right)^3\right]^2\\ \left(x+\dfrac{1}{3}\right)^2=\left(\pm\dfrac{8}{27}\right)^2\)
TH1: \(x+\dfrac{1}{3}=\dfrac{8}{27}\\ x=\dfrac{8}{27}-\dfrac{1}{3}\\ x=-\dfrac{1}{27}\)
TH2:
\(x+\dfrac{1}{3}=-\dfrac{8}{27}\\ x=-\dfrac{8}{27}-\dfrac{1}{3}\\ x=-\dfrac{17}{27}\)
Vậy, `x \in {-1/27; -17/27}.`
(\(x\) + \(\dfrac{1}{3}\))2 = (\(\dfrac{2}{3}\))6
(\(x+\dfrac{1}{3}\))2 = [(\(\dfrac{2}{3}\))3)2
\(\left[{}\begin{matrix}x+\dfrac{1}{3}=\left(\dfrac{2}{3}\right)^3\\x+\dfrac{1}{3}=-\left(\dfrac{2}{3}\right)^3\end{matrix}\right.\)
\(\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{8}{27}\\x+\dfrac{1}{3}=-\dfrac{8}{27}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{8}{27}-\dfrac{1}{3}\\x=-\dfrac{8}{27}-\dfrac{1}{3}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{1}{27}\\x=-\dfrac{17}{27}\end{matrix}\right.\)
Vậy \(x\) \(\in\) { \(\dfrac{-11}{27}\); - \(\dfrac{1}{27}\)}
\(\left(x+\dfrac{1}{3}\right)^2= \left(\dfrac{2}{3}\right)^6\)
\(\Rightarrow x+\dfrac{1}{9}=\dfrac{64}{729}\)
\(\Rightarrow x=\dfrac{64}{729}-\dfrac{1}{9}\)
\(\Rightarrow x=\dfrac{64}{729}-\dfrac{81}{729}\)
\(\Rightarrow x=-\dfrac{17}{729}\)
Vậy...
\(\left(x+\dfrac{1}{3}\right)^2=\left(\dfrac{2}{3}\right)^6\)
\(\Rightarrow\left(x+\dfrac{1}{3}\right)^2=\left[\left(\dfrac{2}{3}\right)^3\right]^2\)
\(\Rightarrow\left(x+\dfrac{1}{3}\right)^2=\left(\pm\dfrac{8}{27}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{8}{27}\\x+\dfrac{1}{3}=-\dfrac{8}{27}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{27}-\dfrac{1}{3}\\x=-\dfrac{8}{27}-\dfrac{1}{3}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{27}\\x=-\dfrac{17}{27}\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{17}{27};-\dfrac{1}{27}\right\}\)
