[(-0.5)3]x = \(\frac{1}{64}\)
<=> \(\left(-\frac{1}{8}\right)^x\)= \(\left(-\frac{1}{8}\right)^2\)
<=> x = 2
\(\left[-\frac{1}{8}\right]^x=\frac{1}{64}\)
\(\frac{-1^x}{8^x}=\frac{1}{8^2}\)
\(\Rightarrow x=2\)
Vậy,........
Ta có \(\left[\left(-0,5\right)^3\right]^x=\frac{1}{64}\)
\(\Rightarrow\left(\left(\frac{-1}{2}\right)^3\right)^x=\frac{1}{64}\)
\(\Rightarrow\left(-\frac{1}{8}\right)^x=\left(-\frac{1}{8}\right)^{-2}\)
\(\Rightarrow x=-2\)