Question

Tinh:

a)\(\sqrt{\sqrt{2019}+\sqrt{2018}}.\sqrt{\sqrt{2019}-\sqrt{2018}}\)

b)\(\sqrt{4+\sqrt{15}}-\sqrt{4-\sqrt{15}}\)

Answer 1

a,

\(\sqrt{\sqrt{2019}+\sqrt{2018}}\cdot\sqrt{\sqrt{2019}-\sqrt{2018}}\\ =\sqrt{\left(\sqrt{2019}+\sqrt{2018}\right)\left(\sqrt{2019}-\sqrt{2018}\right)}\\ =\sqrt{\left(\sqrt{2019}\right)^2-\left(\sqrt{2018}\right)^2}\\ =\sqrt{2019-2018}=\sqrt{1}=1\)

b, Gọi BT cần tìm là A

Ta có:

\(A^2=4+\sqrt{15}+4-\sqrt{15}-2\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}\\ =8-2\sqrt{4^2-\left(\sqrt{15}\right)^2}\\ =8-2\sqrt{16-15}=8-2\cdot1=8-2=6\)

Suy ra \(A=\sqrt{6}\).

Chúc bạn học tốt nha.