biểu thức trên = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=\frac{100}{101}< 1\)
vậy A<1
\(=1-\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}+\frac{1}{101}\)
\(=\left(\frac{1}{1}+\frac{1}{101}\right)\)
\(=\frac{102}{101}\)
\(\Rightarrow A>1\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{3.7}+...+\frac{2}{99.101}>1\)
\(A=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(A=2.\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=1-\frac{1}{101}\)
\(A=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}< 1\left(đpcm\right)\)
