\(\left(-3\right)^2.\left(\dfrac{1}{3}\right)^3:\left[\left(\dfrac{-2}{3}\right)^2+\dfrac{1}{2}-1+\dfrac{1}{3}\right]\)(Có thể mình viết đề sai)
\(=\dfrac{1.1}{3}:\left[\dfrac{4}{9}+\dfrac{1}{2}-1+\dfrac{1}{3}\right]\)
\(=\dfrac{1}{3}:\left[\dfrac{8+9-18+6}{18}\right]\)
\(=\dfrac{1}{3}:\dfrac{5}{18}=\dfrac{1.18}{3.5}\)
\(=\dfrac{1.6}{1.5}=\dfrac{6}{5}\)