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8 tháng 9 2016

em lớp 6  nha

B= 1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42 + 1/56 + 1/72

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

B=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

B=1+0-0-0-0-0-0-0-1/9

B=1-1/9

B=8/9

k em nha

8 tháng 9 2016

phần A em chịu

13 tháng 3 2017

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

= -1/2-1/6-1/12-1/20-1/30-1/42-1/56-1/64-1/72-1/90

= -(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/64+1/72+1/90)

= -(1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10)

= -(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

14 tháng 7 2021

Đây đề lớp 6 mà. Bn này làm đúng rồi nè!

11 tháng 9 2016

 Đặt biểu thức cần tính là A. Ta có : 
A = 9/10 -( 1/90 + 1/72 + ... + 1/2) 
= 9/10 - { 1/( 9.10) + 1/(9.8) + ... + 1/( 2.1)} 
= 9/10 - ( 1/9 - 1/10 + 1/8 - 1/9 + ...+ 1 - 1/2) ( 1/90 = 1/(9.10) = 1/9 - 1/10) 
= 9/10 - ( 1 - 1/10)  

Ta có: \(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\dfrac{8}{9}=0\)

30 tháng 12 2020

ok luôn.hay thì like nha

ta có

1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72

=1/1*2+1/2*3+1/3*4+...+1/8*9

=1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9

=1-1/9

=8/9

9 tháng 2 2021

Ta đặt A=\(-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(\Rightarrow A=-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)\(-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\) 

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

Ta có: \(-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\)

\(=-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}\right)\)

\(=-\left(-\dfrac{1}{10}+1\right)\)

\(=-\left(1-\dfrac{1}{10}\right)\)

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