i) \(\left|x-3\right|=6\Rightarrow\left[{}\begin{matrix}x-3=6\\3-x=6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
k) \(12-\left|x\right|=8\Rightarrow\left|x\right|=4\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
l) \(\left(x-\dfrac{2}{3}\right)^2=\dfrac{5}{6}\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\sqrt{\dfrac{5}{6}}\\x-\dfrac{2}{3}=-\sqrt{\dfrac{5}{6}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{6}}+\dfrac{2}{3}\\x=\dfrac{2}{3}-\sqrt{\dfrac{5}{6}}\end{matrix}\right.\)
m) \(\left(\dfrac{3}{4}-x\right)^3=-8\Rightarrow\dfrac{3}{4}-x=-2\Rightarrow x=\dfrac{11}{4}\)
i: Ta có: \(\left|x-3\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=6\\x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
k: ta có: \(12-\left|x\right|=8\)
\(\Leftrightarrow\left|x\right|=4\)
hay \(x\in\left\{4;-4\right\}\)
