Question

1 Tính 

\(\sqrt{20,8^2-19,2^2}\)

2 Thực hiện phép tính

a) \(2\sqrt{2}.\left(\sqrt{2}-1\right)+\left(1+\sqrt{2}\right)-2\sqrt{6}\)

b) \(\sqrt{2-\sqrt{2}}.\sqrt{2+\sqrt{2}}+8\)

c) \(\left(\sqrt{2-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)

 

Answer 1

2)

a) \(2\sqrt{2}\left(\sqrt{2}-1\right)+\left(1+\sqrt{2}\right)^2-2\sqrt{6}=4-2\sqrt{2}+3+2\sqrt{2}-2\sqrt{6}=7-2\sqrt{6}=\left(\sqrt{6}-1\right)^2\)

b) \(\sqrt{2-\sqrt{2}}.\sqrt{2+\sqrt{2}}+8=\sqrt{2^2-\left(\sqrt{2}\right)^2}=\sqrt{2}+8\)

c) Đề sai.

 

Answer 2

1. \(\sqrt{20,8^2-19,2^2}=\sqrt{\left(20,8-19,2\right)\left(20,8+19,2\right)}=\sqrt{1,6.40}\) 

\(=\sqrt{16.4}=\sqrt{16}.\sqrt{4}=4.2=8\)

Answer 3

2. 

a) \(2\sqrt{2}.\left(\sqrt{2-1}\right)+\left(1+\sqrt{2}\right)-2\sqrt{6}=4-2\sqrt{2}+1+\sqrt{2}-2\sqrt{6}\)

\(=3-\sqrt{2}+2\sqrt{6}\)

b) \(\sqrt{2-\sqrt{2}}.\sqrt{2+\sqrt{2}}+8=\sqrt{4-2}+8=8+\sqrt{2}\)

 

 

Answer 4

mình nhầm đề a rồi

a) \(2\sqrt{2}.\left(\sqrt{2}-1\right)+\left(1+\sqrt{2}\right)^2-2\sqrt{6}\)