Đặt \(\sqrt[3]{2x-1}=m\)
\(\Rightarrow\left\{{}\begin{matrix}x^3=2m-1\\m^3=2x-1\end{matrix}\right.\). Trừ vế theo vế
\(\Rightarrow x^3-m^3=2m-2x\)
\(\Leftrightarrow\left(x-m\right)\left(x^2+xm+m^2\right)=2\left(m-x\right)\)
\(\Leftrightarrow\left(x-m\right)\left(x^2+xm+m^2+2\right)=0\)
\(\Rightarrow x=m\)
\(\Rightarrow\sqrt[3]{2x-1}=x\)
\(\Leftrightarrow x^3=2x-1\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-\dfrac{-1+\sqrt{5}}{2}\right)\left(x-\dfrac{-1-\sqrt{5}}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1+\sqrt{5}}{2}\\x=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\left(n\right)\)
Vậy phương trình có 3 nghiệm . . .
