\(\left(\dfrac{1}{9}\right)^x=\left(\dfrac{1}{27}\right)^6\)
\(\Rightarrow\left(3^{-2}\right)^x=\left(3^{-3}\right)^6\)
\(\Rightarrow3^{-2x}=3^{-18}\)
\(\Rightarrow-2x=-18\)
\(\Rightarrow x=9\)
Vậy \(x=9\)
\(\left(\dfrac{1}{9}\right)^x=\left(\dfrac{1}{27}\right)^6\)
\(3^{-2x}=3^{-18}\)
=>\(-2x=-18\)
=>\(x=9\)
Vậy...