Ta có :
A= \(\dfrac{2\cdot\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
Đặt B=\(2\cdot\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
Ta có B=\(2\cdot\sqrt{3+\sqrt{5-\sqrt{12+2\cdot\sqrt{12}+1}}}\)
\(2\cdot\sqrt{3+\sqrt{5-\sqrt{12}-1}}\\ =2\sqrt{3+\sqrt{4-\sqrt{12}}}\\ =2\cdot\sqrt{3+\sqrt{3-2\cdot\sqrt{3}+1}}\\ =2\cdot\sqrt{3+\sqrt{3}-1}\\ =2\cdot\sqrt{2+\sqrt{3}}\)
Thay B vào A, ta cũng có:
A=\(\dfrac{2\cdot\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\\ =\dfrac{2\cdot\sqrt{2+\sqrt{3}}}{\sqrt{2\cdot\left(\sqrt{3}+1\right)}}\\ =\dfrac{\sqrt{2}\cdot\sqrt{2+\sqrt{3}}}{\sqrt{3}+1}\\ =\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}=\dfrac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}=1\)
Vậy A thuộc Z
