[tex]A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+........+\frac{1}{399\sqrt{400}+400\sqrt{399}}[/tex]
Ta có [TEX]\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}\sqrt{n+1}(\sqrt{n+1}+\sqrt{n})}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt{n})}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}[/TEX]
Áp dụng vào biểu thức [tex]A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+........+\frac{1}{399\sqrt{400}+400\sqrt{399}}[/tex] ta được
[tex]A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+........+\frac{1}{399\sqrt{400}+400\sqrt{399}}[/tex]
[tex]=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}[/tex]
[tex]=1-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}[/tex]
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