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\(\frac{3}{5.11}+\frac{5}{11.21}+\frac{7}{21.35}+\frac{9}{35.53}=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{21}+\frac{1}{21}-\frac{1}{35}+\frac{1}{35}-\frac{1}{53}\right)=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{53}\right)=\frac{1}{10}-\frac{1}{106}<\frac{1}{10}<1\)
\(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{9}{10!}\)
\(A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{10-1}{10!}\)
\(A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{10}{10!}-\frac{1}{10!}\)
\(A=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}+...+\frac{1}{9!}-\frac{1}{10!}\)
\(A=1-\frac{1}{10!}\)
\(\Rightarrow A< 1\left(đpcm\right)\)
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\(\dfrac{3}{5.11}+\dfrac{5}{11.21}+\dfrac{7}{21.35}+\dfrac{9}{35.53}\)
\(=\dfrac{1}{2}\left(\dfrac{6}{5.11}+\dfrac{10}{11.21}+\dfrac{14}{21.35}+\dfrac{18}{35.53}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{53}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{53}\right)< \dfrac{1}{2}.\dfrac{1}{5}=\dfrac{1}{10}< 1\)