\(\left[\left(0,5\right)^3\right]^x=\dfrac{1}{64}\\ \Leftrightarrow\left[\left(\dfrac{1}{2}\right)^3\right]^x=\dfrac{1}{64}\\ \Leftrightarrow\left(\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(\dfrac{1}{8}\right)^2\\ Vậy:x=2\\ \rightarrow S=\left\{2\right\}\)
Ta có: \(\left[\left(-0.5\right)^3\right]^x=\dfrac{1}{64}\)
\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^{3x}=\dfrac{1}{64}\)
\(\Leftrightarrow3x=6\)
hay x=2