A. $6+\sqrt{2}$ ;

B. 6;

C. $3+\sqrt{2}$ ;

D. 9.

Đáp án đúng là: A

Điều kiện xác định x > 2, x ≠ 4

Ta có:${\log }_{\sqrt{3}}\left(x-2\right)+{\log }_3{\left(x-4\right)}^2=0$

$\iff {\log }_3{\left(x-2\right)}^2+{\log }_3{\left(x-4\right)}^2=0$

$\iff {\log }_3{\left(x-2\right)}^2{\left(x-4\right)}^2=0$

$\iff {\left(x-2\right)}^2{\left(x-4\right)}^2=1$

$\iff \left[\begin{array}{l}\left(x-2\right)\left(x-4\right)=1\\ \left(\text{x}-2\right)\left(x-4\right)=-1\end{array}\right.$

$\iff \left[\begin{array}{l}{x}^2-6\text{x + 8}=1\\ x^2-6\text{x}+8=-1\end{array}\right.$

$\iff \left[\begin{array}{l}{x}^2-6\text{x + 7}=0\\ x^2-6\text{x}+9=0\end{array}\right.$

$\iff \left[\begin{array}{l}x=3+\sqrt{2}\\ x=3-\sqrt{2}\\ x=3\end{array}\right.$

Mà x > 2, x ≠ 4

Suy ra $\left[\begin{array}{l}x=3+\sqrt{2}\\ x=3\end{array}\right.$

Vậy tổng các nghiệm là $3+\sqrt{2}+3=6+\sqrt{2}$