Xét biểu thức $A= \sqrt{6,5-\sqrt{12}}+\sqrt{6,5+\sqrt{12}}+2\sqrt6 (A \geq 0)$Bình phương A , ta có
$A^2=6,5.2+(2\sqrt6)^2+2\sqrt{(6,5+\sqrt{12})(6,5-\sqrt{12})}+4\sqrt6(\sqrt{6,5+\sqrt{12}}+\sqrt{6,5-\sqrt{12}})$
$A^2=13+24+2\sqrt{6,5^2-12}+4\sqrt6(\sqrt{6,5-\sqrt{12}}+\sqrt{6,5+\sqrt{12}})$
$A^2=48+4\sqrt6(\sqrt{6,5+\sqrt{12}}+\sqrt{6,5-\sqrt{12}})$
$\Rightarrow (A^2-48)^2=(4\sqrt6)^2(6,5.2+2\sqrt{6,5^2-12})=16.6(13+11)=2304=48^2$
$\Rightarrow A^2-48=48$
$\Rightarrow A^2=96 \Rightarrow A=4\sqrt6 ( do A \geq 0)$