$pt\Leftrightarrow x^3-2x+1=2(\sqrt[3]{2x-1}-x)$$\Leftrightarrow x^3-2x+1=2[\frac{2x-1-x^3}{\sqrt[3]{(2x-1)^2}+\sqrt[3]{2x-1}+1}]$$\Leftrightarrow (x^3-2x+1)(\color{red}{1+\frac{2}{\sqrt[3]{(2x-1)^2}+\sqrt[3]{2x-1}+1}})=0$Dễ thấy biểu thức màu đỏ $>0\Rightarrow x^3-2x+1=0\Rightarrow (x-1)(x^2+x-1)=0$$\Rightarrow x=1,x=\frac{-1\pm\sqrt5}{2}$
$pt\Leftrightarrow x^3-2x+1=2(\sqrt[3]{2x-1}-x)$$\Leftrightarrow x^3-2x+1=2[\frac{2x-1-x^3}{\sqrt[3]{(2x-1)^2}+\sqrt[3]{2x-1}+1}]$$\Leftrightarrow (x^3-2x+1)(\color{red}{1+\frac{2}{\sqrt[3]{(2x-1)^2}+\sqrt[3]{2x-1}+1}})=0$Dễ thấy biểu thức màu đỏ $<0\Rightarrow x^3-2x+1=0\Rightarrow (x-1)(x^2+x-1)=0$$\Rightarrow x=1,x=\frac{-1\pm\sqrt5}{2}$
$pt\Leftrightarrow x^3-2x+1=2(\sqrt[3]{2x-1}-x)$$\Leftrightarrow x^3-2x+1=2[\frac{2x-1-x^3}{\sqrt[3]{(2x-1)^2}+\sqrt[3]{2x-1}+1}]$$\Leftrightarrow (x^3-2x+1)(\color{red}{1+\frac{2}{\sqrt[3]{(2x-1)^2}+\sqrt[3]{2x-1}+1}})=0$Dễ thấy biểu thức màu đỏ $&
gt;0\Rightarrow x^3-2x+1=0\Rightarrow (x-1)(x^2+x-1)=0$$\Rightarrow x=1,x=\frac{-1\pm\sqrt5}{2}$