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\(\frac{1}{1.2}+\frac{1}{3.4}+......+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)\)
\(theocaua\Rightarrow A=\frac{1}{26}+\frac{1}{27}+......+\frac{1}{50}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\left(5sohang\right)+\frac{1}{40}+\frac{1}{40}+....+\frac{1}{40}\left(10sohang\right)+\frac{1}{50}+\frac{1}{50}+....+\frac{1}{50}\left(10sohang\right)=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{35}{60}=\frac{7}{12}\left(1\right)\)
\(A=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}< \frac{1}{25}+\frac{1}{25}+...+\frac{1}{25}\left(5sohang\right)+\frac{1}{30}+\frac{1}{30}+....+\frac{1}{30}\left(10sohang\right)+\frac{1}{40}+\frac{1}{40}+.....+\frac{1}{40}\left(10sohang\right)=\frac{1}{4}+\frac{1}{3}+\frac{1}{5}=\frac{47}{60}< \frac{5}{6}=\frac{50}{60}\left(2\right)\) \(\left(1\right);\left(2\right)\Rightarrow\frac{7}{12}< A< \frac{5}{6}\)
Ta có: 1/1.2 = 1- 1/2
1/3.4 = 1/3 - 1/4
...............
1/19.20 = 1/19 - 1/20
Cộng vế với vế ta đc:
A = 1- 1/20 = 19/20
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{17.18}+\frac{1}{19.20}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{17}-\frac{1}{18}+\frac{1}{19}-\frac{1}{20}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{17}+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+..+\frac{1}{18}+\frac{1}{20}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{18}+\frac{1}{20}\right)\)
\(A=\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}\right)\)
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}\)
\(\frac{A}{B}=1\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
+) \(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+...+\frac{1}{100}\right)>\left(\frac{1}{75}+...+\frac{1}{75}\right)+\left(\frac{1}{100}+...+\frac{1}{100}\right)\)
=> \(A>\frac{25}{75}+\frac{25}{100}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
+) \(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+...+\frac{1}{100}\right)<\left(\frac{1}{50}+...+\frac{1}{50}\right)+\left(\frac{1}{75}+...+\frac{1}{75}\right)\)
=> \(A<\frac{25}{50}+\frac{25}{100}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}<\frac{5}{6}\)
Vậy...
Chứng minh : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\): như câu trên
\(\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}-\frac{19}{9.10}+\frac{21}{10.11}\)
\(=\frac{3+4}{3.4}-\frac{4+5}{4.5}+\frac{5+6}{5.6}-\frac{6+7}{6.7}+\frac{7+8}{7.8}-\frac{8+9}{8.9}-\frac{9+10}{9.10}+\frac{10+11}{10.11}\)
\(=\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}-\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}-\frac{1}{8}-\frac{1}{9}+\frac{1}{9}+\frac{1}{10}-\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{3}-\frac{1}{11}=\frac{8}{33}\)
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{435}\)
\(\Rightarrow A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{870}\)
\(\Rightarrow A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{29.30}\right)\)
\(\Rightarrow A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{29}-\frac{1}{30}\right)\)
\(\Rightarrow A=2\left(\frac{1}{2}-\frac{1}{30}\right)\)
\(\Rightarrow A=1-\frac{1}{15}\)
\(\Rightarrow A=\frac{14}{15}\)
b) \(\frac{\frac{2}{3}+\frac{5}{7}+\frac{4}{21}}{\frac{5}{6}+\frac{11}{7}-\frac{7}{21}}\)
\(=\frac{\frac{29}{21}+\frac{4}{21}}{\frac{101}{42}-\frac{7}{21}}\)
\(=\frac{\frac{11}{7}}{\frac{29}{14}}\)
\(=\frac{22}{29}.\)
Chúc bạn học tốt!