\(A=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right).\left(\sqrt{3}+\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}-\sqrt{3}\right).\left(\sqrt{4}+\sqrt{3}\right)}+...+\frac{\sqrt{121}-\sqrt{120}}{\left(\sqrt{121}-\sqrt{120}\right)\left(\sqrt{121}+\sqrt{120}\right)}\)
\(A=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{121}-\sqrt{120}}{121-120}\)
\(A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{121}-\sqrt{120}\)
\(A=\sqrt{121}-\sqrt{1}=10\)
\(B=\frac{2}{2\sqrt{1}}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{2\sqrt{35}}\)
\(B=2.\left(\frac{1}{\sqrt{1}+\sqrt{1}}+\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+...+\frac{1}{\sqrt{35}+\sqrt{35}}\right)\)
\(>2.\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{35}+\sqrt{36}}\right)\)
\(>2.\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{36}-\sqrt{35}\right)\)
\(=2.\left(\sqrt{36}-\sqrt{1}\right)=2.\left(6-1\right)=10=A\)
Vậy B > A
