\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}\)

\(=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{81}-\sqrt{80}}{81-80}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{81}-\sqrt{80}\)

\(=\sqrt{81}-1=8\)

\(B=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{24}-\sqrt{25}}\)

\(=\frac{\sqrt{1}+\sqrt{2}}{1-2}-\frac{\sqrt{2}+\sqrt{3}}{2-3}+\frac{\sqrt{3}+\sqrt{4}}{3-4}-...-\frac{\sqrt{24}+\sqrt{25}}{24-25}\)

\(=-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{25}+\sqrt{24}\)

\(=\sqrt{25}-\sqrt{1}=4\)

bằng 8

nhanh nha mn

\(\frac{3}{\sqrt{3}-2}+\frac{2}{2+\sqrt{3}}=\frac{3\left(\sqrt{3}+2\right)+2\left(\sqrt{3}-2\right)}{\left(\sqrt{3}-2\right)\left(2+\sqrt{3}\right)}=\frac{3\sqrt{3}+6+2\sqrt{3}-4}{3-4}\)

\(=\frac{5\sqrt{3}+2}{-1}=-5\sqrt{3}-2\)

Ta có: \(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}=\frac{2017-1}{\sqrt{2017}}+\frac{2016+1}{\sqrt{2016}}=\sqrt{2017}+\sqrt{2016}+\left(\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\right)\)

Do \(\sqrt{2016}< \sqrt{2017}\Rightarrow\frac{1}{\sqrt{2016}}>\frac{1}{\sqrt{2017}}\Rightarrow\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}>0\)

=> \(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}>\sqrt{2017}+\sqrt{2016}\)

Hay \(\sqrt{2016}+\sqrt{2017}< \frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}\)

Ta có : \(\frac{2016}{\sqrt{2017}}+\frac{2017}{\sqrt{2016}}\)

\(=\frac{2017-1}{\sqrt{2017}}+\frac{2016+1}{\sqrt{2016}}\)

\(=\sqrt{2016}+\sqrt{2017}+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\)

Vì \(\sqrt{2016}< \sqrt{2017}\)\(\Rightarrow\frac{1}{\sqrt{2016}}>\frac{1}{\sqrt{2017}}\)

\(\Rightarrow\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}>0\)

\(\Rightarrow\sqrt{2016}+\sqrt{2017}+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}>\sqrt{2016}+\sqrt{2017}\)

\(\Rightarrowđpcm\)