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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)
\(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)
\(=\frac{\left[\left(\frac{99}{2}+1\right)+\left(\frac{98}{3}+1\right)+...+\left(\frac{1}{100}+1\right)+\frac{101}{101}\right]}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=\frac{\frac{101}{2}+\frac{101}{3}+...+\frac{101}{100}+\frac{101}{101}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=\frac{101.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=101-2\)( vì \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\ne0\))
\(=99\)
Tham khảo nhé~
Thôi để t làm cho
Ta có \(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\)
= \(100+\frac{101-2}{2}+\frac{101-3}{3}+...+\frac{101-100}{100}\)
= 100 - 99 + \(\frac{101}{2}+\frac{101}{3}+...+\frac{101}{100}\)
= \(1+\frac{101}{2}+\frac{101}{3}+...+\frac{101}{100}\)
= 101(\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\))
Thế vào cái ban đầu được 99
a) \(\frac{53}{101}.\frac{-13}{97}+\frac{53}{101}.\frac{-84}{97}\)
\(=\frac{53}{101}\left(\frac{-13}{97}+\frac{-84}{97}\right)\)
\(=\frac{53}{101}.\frac{-97}{97}\)
\(=\frac{53}{101}.\left(-1\right)\)
\(=\frac{-53}{101}\)
b) \(\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{57}-\frac{1}{5757}\right).0\)
\(=0\)
c) \(\frac{3^2}{25}.\frac{75}{-21}.\frac{50}{35}\)
\(=\frac{3^2.75.50}{25.\left(-21\right).35}\)
\(=\frac{3.3.25.3.5.5.2}{25.3.\left(-7\right).5.7}\)
\(=\frac{3.3.5.2}{\left(-7\right).7}\)
\(=\frac{90}{-49}\)
d) \(\frac{25.48-25.18}{20.5^3}\)
\(=\frac{25\left(48-18\right)}{10.2.125}\)
\(=\frac{25.10.3}{10.2.25.5}\)
\(=\frac{3}{10}\)
ai giúp mk ik
mk đg cần gấp,còn nhìu đề chx lm
