a,Ta có :  \(1-\sqrt{3}\); \(\sqrt{2}-\sqrt{6}=\sqrt{2}\left(1-\sqrt{3}\right)\Rightarrow1-\sqrt{3}< \sqrt{2}\left(1-\sqrt{3}\right)\)

Vậy \(1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)

b, Đặt A =  \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)(*)

\(\sqrt{2}A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)

\(=\sqrt{7}+1-\sqrt{7}+1-2=0\Rightarrow A=0\)

Vậy (*) = 0 

1: 

Ta có: \(\sqrt{2}-\sqrt{6}\)

\(=\sqrt{2}\left(1-\sqrt{3}\right)< 0\)

\(\Leftrightarrow1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)

2:
Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)

\(=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}+1-\sqrt{7}+1-2}{\sqrt{2}}\)

=0

\(-3\sqrt{3}=-\sqrt{27}\)

\(-2\sqrt{7}=-\sqrt{28}\)

mà 27<28

nên \(-3\sqrt{3}>-2\sqrt{7}\)

a: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}=\dfrac{\left(5+2\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}{3}=\dfrac{5\sqrt{5}-5\sqrt{2}+10-2\sqrt{10}}{3}\)

b: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)

viết đề như vậy sao đọc được ?

có ai đọc được ko ???????????

Ta có: \(\sqrt{7}< \sqrt{9}\); \(\sqrt{15}< \sqrt{16}\)

\(\Rightarrow\sqrt{7}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)