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\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)

\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)

\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)

\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)

\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)

\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)

\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)

\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)

\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)

\(=52-\frac{246}{7}\div\frac{82}{63}\)

\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)

\(=52-27=25\)

\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)

\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)

\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)

\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)

\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)

\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)

\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)

21 tháng 6 2018

(1-1/2) x (1-1/3) x .... x (1-1/100) 

= 1/2 x 2/3 x ... x 99/100 

= 1x2x...x99/2x3x..x100 

= 1/ 100 

21 tháng 6 2018

a, \(\frac{3}{5}+25-\frac{1}{5}\)

\(=\left(\frac{3}{5}-\frac{1}{5}\right)+25\)

\(=\frac{2}{5}+25\)

\(=25\frac{2}{5}\)

\(=25,4\)

b, \(13.3.32,27+67,63.39\)

\(=39.32,27+67,63.39\)

\(=39\left(32,37+67,63\right)\)

\(=39.100\)

\(=3900\)

c, \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{100}\right)\)

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

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

21 tháng 12 2016

câu 1 / 2242,82 : 100 + 37411,8 : 1000 = 59,84

câu 2 / (91,5 x 26,3 - 17,4 x 25,6) x ( 41 x 11 - 4100 x 0,1 -41 ) = 0

k với nhé !!!!!!!!!!hihi!!!!!!!!

21 tháng 12 2016

câu 1

2242,82:100 +37411,8:10:100

=2242,82:100+3741,18:100

=(2242,82+3741,18):100

=5984:100

=59,84

17 tháng 1 2019

2

4

6

8

375000

2 tháng 4 2017

Ta có:

\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{100}\right)\)

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

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

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

\(=\frac{2.3.4...99}{3.4.5...100}=\frac{2}{100}=\frac{1}{50}\)

Vậy \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{100}\right)=\frac{1}{50}\)

3 tháng 6 2015

\(=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times...\times\frac{101}{100}=\frac{101}{2}\)

3 tháng 6 2015

=3/2x4/3x5/4....x101/100

=3x4x5x...x101/2x3x4x...x100

=101/2

22 tháng 5 2022

M = \(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{99x100}\)

M = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

M = \(1-\dfrac{1}{100}\)

M = \(\dfrac{99}{100}\)