Gọi \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{17.18}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{17}-\frac{1}{18}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{17}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{18}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{18}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{18}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{18}-1-\frac{1}{2}-\frac{1}{3}-.....-\frac{1}{9}\)
\(=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+....+\frac{1}{18}\)
Ta thấy : \(\frac{1}{10}>\frac{1}{19};\frac{1}{11}>\frac{1}{19};\frac{1}{12}>\frac{1}{19};....;\frac{1}{18}>\frac{1}{19}\)
\(\Rightarrow A=\frac{1}{10}+\frac{1}{11}+...+\frac{1}{18}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}\)(có 9 số \(\frac{1}{19}\) )
\(\Rightarrow A>9.\frac{1}{19}=\frac{9}{19}\)(đpcm)
