Với giá trị nào của $a$ thì phương trình  : ${\log _{\sqrt 3 }}\left( x + 3 \right) = {\log _3}\left(

ax \right)$ có $1$ nghiệm  duy nhất.
Điều kiện: x>-3
$\log_{\sqrt{3}}(x+3)=\log_{3}(ax)\Leftrightarrow \begin{cases}x>-3 \\ (x+3)^2ax \Leftrightarrow x^2+(6-a)x+9=0 \end{cases}         (*)$
Phương trình cho có nghiệm duy nhất nếu và chỉ nếu  $(*)$ có $2$ nghiệm ${x_1},\,{x_2}$ thỏa mãn :
$\left[ \begin{array}{l}
{x_1} = {x_2} >  - 3 &  & \left( a \right)\\
{x_1} \le  - 3 < {x_2} &  & \left( b \right)
\end{array} \right.$
$(a)\Leftrightarrow \begin{cases}\triangle =0 \\ \frac{S}{2}>-3 \end{cases}\\\Leftrightarrow \begin{cases}(6-a)^2-36=0 \\ \frac{a-6}{2}>-3\end{cases}\Leftrightarrow a=12$
Với $a = 12$ ta có ${x_1} = {x_2} = 3$

$(b)\Leftrightarrow \left[ \begin{array}{l}x_{1}=-3<x_{2}\\1.f(-3)<0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}\begin{cases}a=0 \\ x_{1}=x_{2}=-3<-3 (loai)\Leftrightarrow a<0 \end{cases}\\3a<0\end{array} \right.$

Vậy với $a < 0$ hoặc $a = 12$ phương trình có một nghiệm duy nhất.

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