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)

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\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(\Rightarrowđpcm\)

free ire

\(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+...+\frac{1}{100}-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)(ĐPCM)

lon biav đêô

tiik e

\(\frac{1}{51}+\frac{1}{52}+......\frac{1}{100}\)

B= \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\)\(\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)

B= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}\)= \(\frac{1}{20}\)

vậy B= \(\frac{1}{20}\)

b,A=(1/101+1/102+...+1/150)+(1/151+1/152+...1/200)>25/125+25/150+25/175+25/200=(1/5+1/6+1/7)+1/8=107/201+1/8>1/2+2/8=5/8

Vậy A>5/8

Nhớ k mik nha!!!!!!!!!!!!!

a/ Quy đồng mẫu số trong các ngoặc đơn, chúng sẽ giản ước được :\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{18}{19}.\frac{19}{20}=\frac{1}{20}.\) 

b/  Chứng minh   A> 5/8  

\(A=(\frac{1}{101}+...\frac{1}{125})+(\frac{1}{126}+...+\frac{1}{150})+(\frac{1}{151}+...+\frac{1}{175})+\left(\frac{1}{176}+...+\frac{1}{200}\right)\ge.\) 

         \(\ge\frac{25}{125}+\frac{25}{150}+\frac{25}{175}+\frac{25}{200}=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}=\left(\frac{1}{5}+\frac{1}{7}\right)+\left(\frac{1}{6}+\frac{1}{8}\right)=\frac{12}{35}+\frac{7}{24}>\frac{24}{72}+\frac{21}{72}=\frac{45}{72}=\frac{5}{8}\)

a, Ta có: \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=100-\left[1+\left(1-\frac{1}{2}\right)+\left(1-\frac{2}{3}\right)+....+\left(1-\frac{99}{100}\right)\right]\)

\(=100-\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)

\(=100-\left[100-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)

\(=100-100+\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)

\(=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)(đpcm)

b, Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)(đpcm)

a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...\)\(+\frac{99}{100}\)
Xét: \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
    = \(\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)
    = \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)\)                                                          
    = \(\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)( có 99 số hạng là 1 )
    = \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
    = \(\left(99+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
    = \(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrow100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)( đpcm )
Vậy: ... 

a) \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)

\(100=\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(100=1+1+1+...+1\)

\(\Rightarrow100=100\)

b) \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

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) \(\frac{25.48-25.18}{20.5^3}\)

\(=\frac{25\left(48-18\right)}{10.2.125}\)

\(=\frac{25.10.3}{10.2.25.5}\)

\(=\frac{3}{10}\)

 \(\frac{53}{101}.\frac{-13}{97}+\frac{53}{101}.\frac{-84}{97}\)

\(=\frac{53}{101}\left(\frac{-13}{97}+\frac{-84}{97}\right)\)

\(=\frac{53}{101}.\left(-1\right)=\frac{-53}{101}\)