Question

(x+\(\dfrac{1}{3}\))\(^3\)=(\(\dfrac{-1}{8}\))

Answer 1

\(\left(x+\dfrac{1}{3}\right)^3=\dfrac{-1}{8}\)

\(\Leftrightarrow\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{-1}{2}\)

\(\Leftrightarrow x=-\dfrac{1}{6}\)

Vậy .........

Answer 2

\(\left(x+\dfrac{1}{3}\right)^3=-\dfrac{1}{8}\)

=>\(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)

=>\(x+\dfrac{1}{3}=\dfrac{1}{2}\)

=>x=\(\dfrac{1}{2}-\dfrac{1}{3}\)

=>x=\(\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\)

Vậy x=\(\dfrac{1}{6}\)

Answer 3

Ta có: \(\left(x+\dfrac{1}{3}\right)^3=-\dfrac{1}{8}\)

\(\Rightarrow\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)

\(\Rightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\Rightarrow x=-\dfrac{5}{6}\)

Vậy \(x=-\dfrac{5}{6}\)