|
|
a) Đặt
$a=\sqrt{8+2\sqrt{10+2\sqrt{5}}}, b= \sqrt{8-2\sqrt{10+2\sqrt{5}}}$
Ta có
$a^2+b^2=16$
$ab=\sqrt{64-4(10+2\sqrt{5})}=\sqrt{24-8\sqrt{5}}=2\sqrt{6-2\sqrt{5}}=2\sqrt{(\sqrt 5 -1)^2}=2(\sqrt 5 -1)$
Suy ra
$(a-b)^2=a^2+b^2-2ab=16-4(\sqrt 5 -1)=20-4\sqrt 5$
Mà $a>b$ nên ta suy ra
$\sqrt{8+2\sqrt{10+2\sqrt{5}}}- \sqrt{8-2\sqrt{10+2\sqrt{5}}}=a-b=2\sqrt{5-\sqrt 5}$
|