Đáp án: x= $\frac{3+\sqrt[]{5}}{2}$ hoặc x=$\frac{3-\sqrt[]{5}}{2}$
Giải thích các bước giải: ĐKXĐ: $x^{}$≥ $\frac{1}{3}$
$Ta _{}$ $có _{}$ :$x^{2}$ $- x^{}$ $+1^{}$ =2$\sqrt[]{3x-1}$
⇔ $x^{2}$ $+2x^{}$ +$1^{}$= $(\sqrt[]{3x-1}) ^2$ +2$\sqrt[]{3x-1}$ +$1^{}$
⇔ $(x+1)^{2}$ = $(\sqrt[]{3x-1}$$+1^{} )^2$
⇔ \(\left[ \begin{array}{l}\sqrt[]{3x-1}+1^{}=x+1\\\sqrt[]{3x-1}+1^{}=-x-1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}\sqrt[]{3x-1}^{}=x\\\sqrt[]{3x-1}+2^{}=-x\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}{3x-1}^{}=x^{2}\\\sqrt[]{3x-1}+2^{}=-x\end{array} \right.\)
vì $x^{}$≥ $\frac{1}{3}$ nên $x^{2}$ -3x+1=0
⇔ x^2−3x+1=0
⇔ (x-3/2)^2-5/4=0
⇔($x^{}$- $\frac{3}{2}$ - $\frac{\sqrt[]{5}}{2}$ )($x^{}$- $\frac{3}{2}$ + $\frac{\sqrt[]{5}}{2}$ )=0
⇔ $x^{}$= $\frac{3+\sqrt[]{5}}{2}$ hoặc $x^{}$= $\frac{3-\sqrt[]{5}}{2}$ (thỏa mãn ĐKXĐ)
