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23 tháng 7 2017

\(C=\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(C=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.8}+\frac{1}{9.10}\right)\)

\(C=\frac{9}{10}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

\(C=\frac{9}{10}-\left(\frac{1}{1}-\frac{1}{10}\right)\)

\(C=\frac{9}{10}-\frac{9}{10}=0\)

14 tháng 4 2018

1 1/9 x 1 1/10 x 1 1/11 x ... x 1 1/2011

=10/9 x 11/10 x 12/11 x ... x 2012/2011

khử

còn 2012/9

13 tháng 4 2018

=\(\frac{10}{9}\)x\(\frac{11}{10}\)x\(\frac{12}{11}\)x.........x\(\frac{2012}{2011}\)

=\(\frac{2012}{9}\)

3 tháng 6 2019

#)Giải :

\(\left(92-\frac{1}{9}-\frac{2}{10}-\frac{3}{10}-...-\frac{92}{100}\right):\left(\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}\right)\)

\(=\left(1-\frac{1}{9}+1-\frac{2}{10}+1-\frac{3}{11}+...+1-\frac{92}{100}\right)\div\frac{1}{5}\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)

\(=\left(\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}\right)\div\frac{1}{5}\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)

\(=8\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\div\frac{1}{5}\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)

\(=8\div\frac{1}{5}\)

\(=40\)

                         #~Will~be~Pens~#

12 tháng 6 2018

\(a,\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}.\)

\(b,\left[x\cdot\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}:\frac{9}{4}\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}\cdot\frac{4}{9}\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{14}{9}\)

\(\Leftrightarrow x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)

\(\Leftrightarrow x\cdot\frac{5}{3}-1=14\)

\(\Leftrightarrow x\cdot\frac{5}{3}=14+1\)

\(\Leftrightarrow x\cdot\frac{5}{3}=15\)

\(\Leftrightarrow x=15:\frac{5}{3}\)

\(\Leftrightarrow x=15\cdot\frac{3}{5}\)

\(\Leftrightarrow x=9.\)

12 tháng 6 2018

a)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=\frac{1}{1}-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

b)\(\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)

\(\Leftrightarrow\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:\frac{9}{4}=1\frac{5}{9}\)

\(\Rightarrow x.\frac{5}{3}-1=1\frac{5}{9}.9=14\)

\(\Rightarrow x.\frac{5}{3}=14+1=15\)

\(\Rightarrow x=15:\frac{5}{3}=9\)

4 tháng 8 2018

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

\(=1-\frac{1}{1000}+1\)

\(=\frac{1000}{1000}-\frac{1}{1000}+\frac{1000}{1000}\)

\(=\frac{1999}{1000}\)

Tham khảo nhé~

4 tháng 8 2018

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)

\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

\(1-\frac{1}{1000}+1\)

\(\frac{1999}{1000}\)

16 tháng 4 2021

Đây là toán lớp 6 nha

1 tháng 7 2016

1/1x2+1/2x3+1/3x4+...+1/99x100+1

1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100 +1

=1- 1/100 +1

=99/100 +1

=199/100

1 tháng 7 2016

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/999.1000 + 1

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/999 - 1/1000 + 1

= 1 - 1/1000 + 1

= 2 - 1/1000

= 2000/1000 - 1/1000

= 1999/1000

Ủng hộ mk nha ♡_♡☆_☆

18 tháng 8 2023

 \(\dfrac{3}{4}\times\dfrac{8}{5}:1\dfrac{1}{6}\) 

=\(\dfrac{6}{5}:\) \(\dfrac{7}{6}\) 

=\(\dfrac{6}{5}\times\dfrac{6}{7}=\dfrac{36}{35}\)

 

2\(\dfrac{1}{3}\) x 1\(\dfrac{1}{4}\) -\(\dfrac{7}{5}\)

\(\dfrac{7}{3}\times\dfrac{5}{4}-\) \(\dfrac{7}{5}\) 

\(\dfrac{35}{12}-\dfrac{7}{5}\)

\(\dfrac{175}{60}-\dfrac{84}{60}=\dfrac{91}{60}\)

 

18 tháng 8 2023

4\(\dfrac{2}{3}+1\dfrac{1}{4} +2\dfrac{1}{3}+2\dfrac{3}{7}\)

(4 +2) + \(\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\) +1\(\dfrac{1}{4}\) + \(2\dfrac{3}{7}\) 

6 + 1 + \(\dfrac{5}{4}\) + \(\dfrac{17}{7}\)

7 + \(\dfrac{103}{28}\)

\(\dfrac{299}{28}\)