\(\left(3x-1\right)^3=-\dfrac{8}{27}\\ \Leftrightarrow3x-1=-\dfrac{2}{3}\\ \Leftrightarrow3x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{9}\)
\(\Leftrightarrow3x-1=-\dfrac{2}{3}\)
hay x=1/9
\(\left(3x-1\right)^3=-\left(\dfrac{2}{3}\right)^3\\ 3x-1=-\dfrac{2}{3}\\ 3x=-\dfrac{2}{3}+1\\ 3x=\dfrac{5}{3}\\ x=\dfrac{5}{3}:3\\ x=\dfrac{5}{9}\)
Ta có :
\(\left(3x-1\right)^3=\dfrac{-8}{27}\)
\(\left(3x-1\right)^3=\left(\dfrac{-2}{3}\right)^3\)
\(=>3x-1=\dfrac{-2}{3}\)
\(=>3x=1+\dfrac{-2}{3}\)
\(=>3x=\dfrac{1}{3}\)
\(=>x=\dfrac{1}{9}\)
