Đáp án:$ĐKXĐ:x>3$
$2\sqrt{x^2-9} =(x+5)\sqrt{\frac{x+3}{x-3}}$
$2\sqrt{x^2-9}-(x+5)\sqrt{\frac{x+3}{x-3}} =0$
$2\sqrt{(x-3)(x+3)}-(x+5)\sqrt{\frac{x+3}{x-3}} =0$
$\sqrt{(x+3)}(2\sqrt{x-3}-(x+5)\sqrt{\frac{1}{x-3}})=0$
$\left \{ {{\sqrt{x+3}=0(loại )} \atop {2\sqrt{x-3}-(x+5)\sqrt{\frac{1}{x-3}}=0}} \right.$
$⇒2(x-3)=(x+5)$
$⇔2x-6-x-5=0$
$⇔x=11$ (t/m)
Vậy có nghiệm x=11