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ta có:\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
=\(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\) \(-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)\)
=\(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-\) \(2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
=\(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
Chứng minh rằng \(\frac{7}{12}<\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{40}<\frac{5}{6}\)
VÌ 1/1.1/3.......1/99=2/51.2/52.........2/100
VÀ 2/51.2/52.....2/100=1/1.1/3.......1/99
SUY RA BẰNG NHAU
Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)(đpcm)
RÚT GỌN:
\(\frac{3.13-13.18}{15.40-80}=\frac{13.\left(3-18\right)}{15.40-40.2}=\frac{13.\left(-15\right)}{40\left(15-2\right)}=\frac{13.\left(-15\right)}{40.13}=-\frac{15}{40}=-\frac{3}{8}\)
CHỨNG MINH:
Ta thấy \(\frac{1}{101}>\frac{1}{150};\frac{1}{102}>\frac{1}{150};...;\frac{1}{149}>\frac{1}{150}\)
=>\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}>\frac{1}{150}.\left(150-101+1\right)=\frac{1}{150}.50=\frac{50}{150}=\frac{1}{3}\)(đpcm)
TÍNH HỢP LÝ:
B=\(\frac{5}{13}+\frac{-5}{7}-\frac{20}{41}+\frac{8}{13}+\frac{-21}{41}=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-21}{41}-\frac{20}{41}\right)+\frac{-5}{7}=1+\left(-1\right)+\frac{-5}{7}=0+\frac{-5}{7}=\frac{-5}{7}\)
Ta có :
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)