Đáp án:

$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
{x^2} - 9 \ge 0\\
{x^2} - 6x + 9 \ge 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
\left( {x - 3} \right)\left( {x + 3} \right) \ge 0\\
{\left( {x - 3} \right)^2} \ge 0\left( {\text{luôn đúng}} \right)
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x \ge 3\\
x \le  - 3
\end{array} \right.\\
\sqrt {{x^2} - 9} .\sqrt {{x^2} - 6x + 9}  = 0\\
 \Rightarrow \left[ \begin{array}{l}
{x^2} - 9 = 0\\
{x^2} - 6x + 9 = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
{x^2} = 9\\
{\left( {x - 3} \right)^2} = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
x =  - 3\left( {tm} \right)\\
x = 3\left( {tm} \right)
\end{array} \right.\\
\text{Vậy}\,x = 3\,\text{hoặc}\,x =  - 3
\end{array}$

Đáp án:

Giải thích các bước giải:

 $\begin{array}{l} Dkxd:\left\{ \begin{array}{l} {x^2} - 9 \ge 0\\ {x^2} - 6x + 9 \ge 0 \end{array} \right.\\  \Rightarrow \left\{ \begin{array}{l} \left( {x - 3} \right)\left( {x + 3} \right) \ge 0\\ {\left( {x - 3} \right)^2} \ge 0\left( {\text{luôn đúng}} \right) \end{array} \right.\\  \Rightarrow \left[ \begin{array}{l} x \ge 3\\ x \le  - 3 \end{array} \right.\\ \sqrt {{x^2} - 9} .\sqrt {{x^2} - 6x + 9}  = 0\\  \Rightarrow \left[ \begin{array}{l} {x^2} - 9 = 0\\ {x^2} - 6x + 9 = 0 \end{array} \right.\\  \Rightarrow \left[ \begin{array}{l} {x^2} = 9\\ {\left( {x - 3} \right)^2} = 0 \end{array} \right.\\  \Rightarrow \left[ \begin{array}{l} x = 3\left( {tm} \right)\\ x =  - 3\left( {tm} \right)\\ x = 3\left( {tm} \right) \end{array} \right.\\ \text{Vậy}\,x = 3\,\text{hoặc}\,x =  - 3 \end{array}$