Xét hiệu :

\(A-B=2\left(\sqrt{1}-\sqrt{2}\right)+2.\left(\sqrt{3}-\sqrt{4}\right)+...+2\left(\sqrt{19}-\sqrt{20}\right)\)

Mà: \(\sqrt{1}

Ta có : \(\sqrt{3}-\sqrt{2}=\dfrac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=\dfrac{1}{\sqrt{3}+\sqrt{2}}>\dfrac{1}{\sqrt{4}+\sqrt{3}}=\sqrt{4}-\sqrt{3}\Rightarrow\sqrt{3}-\sqrt{2}>\sqrt{4}-\sqrt{3}\Rightarrow2\sqrt{3}>\sqrt{4}+\sqrt{2}\)

Làm tương tự : \(2\sqrt{5}>\sqrt{4}+\sqrt{6};2\sqrt{7}>\sqrt{6}+\sqrt{8},...,2\sqrt{19}>\sqrt{18}+\sqrt{20}\)

Cộng từng BĐT trên , ta được :

\(2\sqrt{3}+2\sqrt{5}+...+2\sqrt{19}>\sqrt{4}+\sqrt{2}+\sqrt{4}+\sqrt{6}+...+\sqrt{18}+\sqrt{20}=2\sqrt{4}+2\sqrt{6}+...+2\sqrt{18}+\sqrt{20}+\sqrt{2}\)

\(\Leftrightarrow A-2\sqrt{1}>B-\sqrt{2}\)

\(\Leftrightarrow A-B>2-\sqrt{2}>0\Rightarrow A>B\)

\(a,=2\sqrt{6}-4+\sqrt{\left(3-\sqrt{6}\right)^2}=2\sqrt{6}-4+3-\sqrt{6}=\sqrt{6}-1\\ b,=3-2\sqrt{2}+\sqrt{\left(3\sqrt{2}+1\right)^2}=3-2\sqrt{2}+3\sqrt{2}+1=4+\sqrt{2}\\ c,=\sqrt{\left(\sqrt{5}+2\right)^2}-\left(\sqrt{5}-1\right)=\sqrt{5}+2-\sqrt{5}+1=3\)

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

b) \(=\left|3-2\sqrt{2}\right|+\sqrt{\left(3\sqrt{2}+1\right)^2}=3-2\sqrt{2}+3\sqrt{2}+1=4+\sqrt{2}\)

c) \(=\sqrt{\left(\sqrt{5}+2\right)^2}-\left|1-\sqrt{5}\right|=\sqrt{5}+2+1-\sqrt{5}=3\)

a)\(\sqrt{\left(2\sqrt{6}-4\right)^2}+\sqrt{\left(3-\sqrt{6}\right)^2}\)=2\(\sqrt{6}-4+3-\sqrt{6}\)=\(\sqrt{6}-1\)
b)\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(1+\sqrt{18}\right)^2}\)=3-2\(\sqrt{2}+1+3\sqrt{2}\)=4+\(\sqrt{2}\)
c)\(\sqrt{\left(2+\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}\)=2+\(\sqrt{5}+1-\sqrt{5}\)=3

Mình cần trình bày ạ!