√(x-3) -2 √(x^2-9) =0

`sqrt(x-3) - 2sqrt(x^2-9) =0`     (ĐK: `x>=3`)

`<=> sqrt(x-3) - 2. sqrt((x-3)(x+3))=0`

`<=> sqrt(x-3) - 2. sqrt(x-3). sqrt(x+3) = 0`

`<=> sqrt(x-3)(1-2sqrt(x+3))=0`

`<=>` \(\left[ \begin{array}{l}√(x-3)=0\\1-2√(x+3)=0\end{array} \right.\)  

`<=>` \(\left[ \begin{array}{l}x=3\\2√(x+3)=1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=3\\√(x+3)=1/2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=3\\x+3=1/4\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=3\\x=-11/4 (ko thoả mãn)\end{array} \right.\) 

Vậy `x=3`