\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}=\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+...++\frac{1}{299}+\frac{1}{300}\right)\)
\(=\left(\frac{1}{200}.100\right)+\left(\frac{1}{300}.100\right)\)
\(=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}>\frac{4}{6}=\frac{2}{3}\)
\(Vậy\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\RightarrowĐPCM\)
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