căn x bình -9 = 3 căn x -3

#andy

\[\begin{array}{l}
\sqrt {{x^2} - 9}  = 3\sqrt {x - 3} \,\,DK:x \ge 3\\
 \Leftrightarrow \sqrt {(x + 3)(x - 3)}  - 3\sqrt {x - 3}  = 0\\
 \Leftrightarrow \sqrt {x - 3} (\sqrt {x + 3}  - 3) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 3}  = 0\\
\sqrt {x + 3}  - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 6
\end{array} \right.\\
 \Rightarrow S \in \{ 3;6\} 
\end{array}\]

$\sqrt[]{x^2-9}=3\sqrt[]{x-3}$  $ĐKXĐ:x≥3$

$<=>\sqrt[]{x-3}.\sqrt[]{x+3}-3\sqrt[]{x-3}=0$

$<=>\sqrt[]{x-3}(\sqrt[]{x+3}-3)=0$

$<=>\left[\begin{matrix} \sqrt[]{x-3}=0\\ \sqrt[]{x+3}-3=0\end{matrix}\right.$

$<=>\left[\begin{matrix} x-3=0\\ \sqrt[]{x+3}=3\end{matrix}\right.$

$<=>\left[\begin{matrix} x=3\\ x+3=9\end{matrix}\right.$

$<=>\left[\begin{matrix} x=3,tm\\ x=6,tm\end{matrix}\right.$

Vậy $x=3$ hoặc $x=6$