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2 tháng 4 2023

1+1=3 :)))

5 tháng 6 2015

\(\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{10}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)\(=\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\frac{1}{100}\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)\(=\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot0\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)=0

5 tháng 6 2015

\(\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right).\left(\frac{1}{100}-\left(\frac{1}{2}\right)^2\right)......\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)

\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right)....\left(\frac{1}{100}-\left(\frac{1}{10}\right)^2\right)...\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)

\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right)...\left(\frac{1}{100}-\frac{1}{100}\right)...\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)

\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right).....0......\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)

\(=0\)

 

27 tháng 5 2017

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)

27 tháng 5 2017

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: ... 

12 tháng 7 2016

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

\(=\frac{3}{2}.\frac{4}{3}.........\frac{100}{99}\)

\(=\frac{100}{2}=50\)

\(B=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).........\left(\frac{1}{100}-1\right)\)

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

\(=\frac{-1}{100}\)

12 tháng 7 2016

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

  \(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

\(=\frac{3.4.5.....100}{2.3.4.....99}\)

 \(=\frac{100}{2}=50\)

11 tháng 3 2018

A= (1/4-1)(1/9-1).......(1/10000-1)

A=-3/4(-8/9).........(-9999/100^2)

A=-1.3/2.2 (-2.4/3.3)........(-99.101/100.100)

A=-1.(-2).(-3)........(-99)/2.3.4......100   .    2.3.4......101/.3.4....100

A=-1/100 . 102/3=17/50

Vậy A= 17/50

12 tháng 3 2018

làm sao để viết được p/s và số mũ như bạn vậy?