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Ta có:\(\frac{1}{2^2}=\frac{1}{4};\frac{1}{3^2}< \frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3};\frac{1}{3^2}< \frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4};.....;\frac{1}{100^2}< \frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{3}{4}\left(đpcm\right)\)
Gọi \(D=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}< \frac{3}{4}\)
Vì \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{100^2}< \frac{1}{99.100}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< \frac{3}{4}\)
\(\Rightarrow D< \frac{3}{4}\left(đpcm\right)\)
1/ Tính:
\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
\(=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{19}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+........+\frac{1}{1317}-\frac{1}{1318}+\frac{1}{1319}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{1319}-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{1318}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{1319}-\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{658}+\frac{1}{659}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{659}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{659}\right)+\frac{1}{660}+\frac{1}{661}+......+\frac{1}{1319}\)
\(=\frac{1}{660}+\frac{1}{661}+.........+\frac{1}{1319}\)