+ ĐKXĐ : \(5x^3\ge1\)
\(pt\Leftrightarrow\left(\sqrt{5x^3-1}-2\right)+\left(\sqrt[3]{2x-1}-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\frac{5x^3-1-4}{\sqrt{5x^3-1}+2}+\frac{2x-1-1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1}+\left(x-1\right)=0\)
\(\Leftrightarrow\frac{5\left(x-1\right)\left(x^2+x+1\right)}{\sqrt{5x^3-1}+2}+\frac{2\left(x-1\right)}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1}+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\frac{5\left(x^2+x+1\right)}{\sqrt{5x^3-1}+2}+\frac{2}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1}+1\right]=0\) (1)
Vì \(\frac{5\left(x^2+x+1\right)}{\sqrt{5x^3-1}+2}+\frac{2}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1}+1>0\forall xTMĐK\) nên từ (1) suy ra :
\(x-1=0\Leftrightarrow x=1\) ( TM )
