\(A^2=3940+2\cdot\sqrt{1970^2-1}\)

\(B^2=3940+2\cdot\sqrt{1970^2}\)

mà \(1970^2-1< 1970^2\)

nên A<B

\(a=\sqrt{1969}+\sqrt{1971}\)

\(\Rightarrow a^2=1969+2\sqrt{1969\cdot1971}+1971\)

\(\Rightarrow a^2=2\cdot1970+2\sqrt{1969\cdot1971}\)                        (1)

\(b=2\cdot\sqrt{1970}\)

\(\Rightarrow b^2=4\cdot1970=2\cdot1970+2\cdot1970\)                   (2)

có : \(1969+1971\ge2\sqrt{1969\cdot1971}\)

\(\Rightarrow2\cdot1970\ge2\sqrt{1969\cdot1971}\)    vì 1969 khác 1971

\(\Rightarrow2\cdot1970>2\sqrt{1969\cdot1971}\)               (3)

\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow a^2< b^2\) mà a;b không âm

\(\Rightarrow a< b\)

Bình phương a và b lên để so sánh

kết quả là a=b nha bạn

Kết quả là a = b đó

b lớn hơn nha bạn 

\(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>2^2=4\left(5>4\right)\\ \Leftrightarrow\sqrt{2}+\sqrt{3}>2\)

\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40};\left(\sqrt{7}-\sqrt{6}\right)^2=13-2\sqrt{42}\\ 2\sqrt{40}>0>-2\sqrt{42}\\ \Leftrightarrow13+2\sqrt{40}>13-2\sqrt{42}\\ \Leftrightarrow\left(\sqrt{8}+\sqrt{5}\right)^2>\left(\sqrt{7}-\sqrt{6}\right)^2\\ \Leftrightarrow\sqrt{8}+\sqrt{5}>\sqrt{7}-\sqrt{6}\)

\(\sqrt{2}\) + \(\sqrt{3}\)  > 2

\(a,\left(\sqrt{2}+\sqrt{11}\right)^2=12+2\sqrt{22}\\ \left(\sqrt{3}+5\right)^2=28+10\sqrt{3}\)

Ta thấy \(12< 28;2\sqrt{22}=\sqrt{88}< \sqrt{300}=10\sqrt{3}\)

Nên \(\sqrt{2}+\sqrt{11}< \sqrt{3}+5\)

\(b,\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\\ \left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)

Vì \(\sqrt{105}< \sqrt{120}\Rightarrow-2\sqrt{105}>-2\sqrt{120}\)

Nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)