Ta có:
\(\left(-\frac{3}{4}\right)^{3x-1}=-\frac{27}{64}\Rightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)
\(\Rightarrow3x-1=3\Rightarrow3x=4\Rightarrow x=\frac{4}{3}\)
Ta có : \(\left(\frac{-3}{4}\right)^{3x-1}=\frac{-27}{64}\)
⇔ \(\left(\frac{-3}{4}\right)^{3x-1}=\left(\frac{-3}{4}\right)^3\) ⇔ \(3x-1=3\) ⇔ \(x=\left(3+1\right):3\) ⇔ \(x=\frac{4}{3}\)
\(\left(-\frac{3}{4}\right)^{3x-1}=-\frac{27}{64}\)
\(\Rightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)
\(\Rightarrow3x-1=3\)
\(\Rightarrow3x=3+1\)
\(\Rightarrow3x=4\)
\(\Rightarrow x=4:3\)
\(\Rightarrow x=\frac{4}{3}\)
Vậy \(x=\frac{4}{3}.\)
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