Ta có: \(\left(2x+\frac{1}{3}\right)^3=\frac{-8}{27}\)
\(\Leftrightarrow2x+\frac{1}{3}=\frac{-2}{3}\)
\(\Leftrightarrow2x=\frac{-2}{3}-\frac{1}{3}=-1\)
hay \(x=-\frac{1}{2}\)
Vậy: \(x=-\frac{1}{2}\)
\(\left(2x+\frac{1}{3}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(2x+\frac{1}{3}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow2x+\frac{1}{3}=-\frac{2}{3}\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}\)
Ta có:\((2x+\frac{1}{3})^3=\frac{-8}{27}\)
\(\Rightarrow2x+\frac{1}{3}=-\frac{2}{3}\)
\(\Rightarrow2x=-\frac{2}{3}-\frac{1}{3}\)
\(\Rightarrow2x=-1\) hay \(x=-\frac{1}{2}\)
Vậy x=\(-\frac{1}{2}\)
Nhớ tick cho mình nhé!
