$x$-$\sqrt[]{6}$$\sqrt{x}$$+1=0$
⇔$x$-$\sqrt[]{6x}$$+1=0$
⇔$-$$\sqrt{6x}$ =$-x-1$
⇔$\sqrt{6x}$$=x+1$
⇔$6x$=$x^{2}$$+2x+1$
⇔$6x-$$x^{2}$$-2x-1=0$
⇔$4x-$$x^{2}$ $-1=0$
⇔$-$$x^{2}$ $+4x-1=0$
⇔$x^{2}$ $-4x+1=0$
⇔$(2+$$\sqrt[]{3}$$)$$(2-$$\sqrt{3}$$)=0$
⇔\(\left[ \begin{array}{l}x=2-\sqrt{3}\\x=2+√3\end{array} \right.\)