Giải thích các bước giải:
$\dfrac{\sqrt{2x+2\sqrt{x^2-9}}}{\sqrt{x^2-9}+x+3}$
$=\dfrac{\sqrt{x-3+2\sqrt{(x-3)(x+3)}+x+3}}{\sqrt{(x-3)(x+3)}+x+3}$
$=\dfrac{\sqrt{(\sqrt{x-3}+\sqrt{x+3})^2}}{\sqrt{x+3}(\sqrt{x-3}+\sqrt{x+3})}$
$=\dfrac{\sqrt{x-3}+\sqrt{x+3}}{\sqrt{x+3}(\sqrt{x-3}+\sqrt{x+3})}$
$=\dfrac{1}{\sqrt{x+3}}$
Mà $x=2\sqrt{6}+2\rightarrow x+3=3+2\sqrt{6}+2=3+2.\sqrt{3}.\sqrt{2}+2=(\sqrt{3}+\sqrt{2})^2$
$\rightarrow\sqrt{x+3}=\sqrt{3}+\sqrt{2}$
$\rightarrow $$\dfrac{\sqrt{2x+2\sqrt{x^2-9}}}{\sqrt{x^2-9}+x+3}=\dfrac{1}{\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\dfrac{\sqrt{3}-\sqrt{2}}{3-2}=\sqrt{3}-\sqrt{2}$
