$\sqrt[3]{2x-1}+\sqrt[3]{x-1}=1$
$<=>\sqrt[3]{x-1+x-1+1}+\sqrt[3]{x-1}=1 $
đặt `t=x-1(1)`
`<=>`$\sqrt[3]{2t+1}+\sqrt[3]{t}=1 $
`<=>`$(\sqrt[3]{2t+1})^3=(1-\sqrt[3]{t})^3 $
`<=>`$2t+1=1-3\sqrt[3]{t}+3(\sqrt[3]{t})^2-t=0$
`<=>`$3t+3\sqrt[3]{t}-3( \sqrt[3]{t})^2=0$
`<=>`$t-(\sqrt[3]{t})^2+\sqrt[3]{t}=0 $
đặt $m=\sqrt[3]{t}$
`<=>m^3-m^2+m=0`
`<=>m^3+0-m^2+0+m=0`
`<=>m^2(m+0)-m(m+0)+m+0=0`
`<=>(m^2-m+1)(m-0)=0`
`<=>`\(\left[ \begin{array}{l}m^2-m+1=0\\m-0=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}m^2-2.\dfrac{1}{2 }+\dfrac{1}{4 }+3=0\\m-0=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}(m-\dfrac{1}{2 })+3=0\\m-0=0\end{array} \right.\)
`<=>m=0` vì `(m-1/2)+3`$\neq0$
với `m=0=>t=0`
thay `t=0` vào `(1)` ta có:
`x-1=0`
`<=>x=1`
vậy `x=1`
