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21 tháng 3 2020

\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)

\(A=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{99\cdot100}\)

\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{4}-\frac{1}{100}=\frac{25}{100}-\frac{1}{100}=\frac{24}{100}=\frac{6}{25}\)

28 tháng 8 2020

\(=\frac{8}{9}+\frac{1}{2}-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=\frac{8}{9}+\frac{1}{2}-\left(\frac{1}{3}-\frac{1}{9}\right)=1+\frac{1}{2}-\frac{1}{3}=1\frac{1}{6}\)

3 tháng 8 2019

\(B=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-...-\frac{1}{6}-\frac{1}{2}\)

\(-B=\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+...+\frac{1}{6}+\frac{1}{2}\)

\(-B=\frac{1}{10.9}+\frac{1}{9.8}+\frac{1}{8.7}+...+\frac{1}{3.2}+\frac{1}{2.1}\)

\(-B=\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+...+\frac{1}{2}-1\)

\(-B=\frac{1}{10}-1\)

\(-B=\frac{9}{10}\)

=> \(B=\frac{-9}{10}\)

3 tháng 8 2019

\(B=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{1}{90}-\left(\frac{1}{72}+\frac{1}{56}+...+\frac{1}{6}+\frac{1}{2}\right)\)

\(=\frac{1}{90}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}+\frac{1}{72}\right)\)

\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\frac{8}{9}\)

\(=-\frac{79}{90}\)

10 tháng 4 2018

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)

\(\Rightarrow B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

           \(=1-\frac{1}{100}\)

            \(=\frac{99}{100}\)

21 tháng 5 2019

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

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

\(B=1-\frac{1}{100}=\frac{99}{100}\)

~ Hok tốt ~

5 tháng 10 2019

Đặt :
\(\frac{1}{315}=a;\frac{1}{651}=b\)  thay vào A ta được :

\(A=\left(2+a\right)b-\left(3+1-b\right).3a-4ab+12a\)

\(\Leftrightarrow A=2b+ab-12a+3ab-4ab+12a\)

\(\Leftrightarrow A=2b\)

Thay \(b=\frac{1}{651}\) ta được :

\(A=\frac{2}{651}\)

Chúc bạn học tốt !!!

17 tháng 4 2017

A = \(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)\(\frac{1}{3}-\frac{1}{10}=\frac{7}{30}\)

17 tháng 4 2017

\(A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\)

\(A=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)\

\(A=\frac{1}{3}-\frac{1}{10}\)

\(A=\frac{7}{30}\)

19 tháng 4 2017

B = 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90

B = \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

B = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\)\(\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

B = \(\frac{1}{2}-\frac{1}{10}\)

B = \(\frac{2}{5}\)

19 tháng 4 2017

B=1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90

B=1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10

B=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10

B=1/2-1/10

B=2/5

8 tháng 8 2016

a) 1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12  - 1/6 - 1/2

 = 1/10.9 - 1/9.8 - 1/8.7 - 1/7.6 - 1/6.5 - 1/5.4 - 1/4.3 - 1/3.2 - 1/2.1

=  1/10 - 1

=   0,1 - 1

=      -0,9

5 tháng 4 2017

C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)

c=\(\frac{1}{1}-\frac{1}{10}\)

c=\(\frac{9}{10}\)

còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!

12 tháng 3 2016

= - ( 1/2 +1/6+1/12+1/20+ 1/30+ 1/42+ 1/56+ 1/72+ 1/90)

= - ( 1 - 1/2 + 1/2 -1/3 +1/3 -1/4 +1/4 - 1/5 +1/5 -1/6 +1/6 -1/7 +1/7 -1/8 +1/8 -1/9 +1/9 -1/10)

= - ( 1- 1/10 )

= -9/10

\(-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\left(-\frac{1}{90}\right)+\left(-\frac{1}{72}\right)+\left(-\frac{1}{56}\right)+......+\left(-\frac{1}{2}\right)\)

\(=\left(-1\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{9.10}\right)\)

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

\(=\left(-1\right).\left(1-\frac{1}{90}\right)=\frac{\left(-1\right).89}{90}=-\frac{89}{90}\)

nha