\(\left[\left(-0,5\right)^3\right]^x=\dfrac{1}{64}\)
\(\Rightarrow\left(-\dfrac{1}{8}\right)^x=\left(-\dfrac{1}{8}\right)^2\)
\(\Rightarrow x=2\)
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