\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

\(C=\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{31\times34}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{31}-\frac{1}{34}\)

\(=1-\frac{1}{34}=\frac{33}{34}\)

99/34

\(C=3-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+\frac{3}{7}-\frac{3}{10}+...+\frac{3}{31}-\frac{3}{34}\)

\(C=3+\frac{3}{34}=\frac{99}{34}\)

:V Làm sai hết rồi sai ngay từ bước đầu tiên.

\(\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-....-\frac{1}{9.10}\)

\(=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}\right)\)

\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{10}\right)\)

\(=\frac{1}{12}-\frac{3}{20}\)

\(=\frac{-11}{12}\)

\(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{9.10}\)

= \(-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

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

= \(-\left(\frac{1}{3}-\frac{1}{10}\right)\)

= \(-\frac{7}{30}\)

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

\(=-\left(\frac{1}{3}-\frac{1}{10}\right)\)

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

~ Hok tốt ~

   1 x 3 x 5 + 2 x 6 x 10 + 4 x 10 x 12 + 7 x 21 x 35 / 1 x 5 x 7 + 2 x 10 x 14 + 4 x 20 x 28 + 7 x 35 x 49

= 3 / 7 + 6 / 14 + 10 x 2 x 6 / 10 x 2 x 28 + 21 / 49

= 3 / 7 + 6 / 14 + 6 / 28 + 21 / 49 

= 3 / 7 + 6 / 14 + 3 / 14 + 3 / 7 

= ( 3 / 7 + 3 / 7 ) + ( 6 / 14 + 3 / 14 )

=         6 / 7 + 9 14 

= 12 / 14 + 9 / 14 

= 21 / 14

nho hem 

lm dau tien lun do 

dung 100% nha 

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+......+\frac{1}{x\left(x+3\right)}=\frac{6}{19}\)

\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.......+\frac{1}{x}-\frac{1}{x+3}=\frac{6}{19}\)

\(1-\frac{1}{x+3}=\frac{6}{19}\)

\(\frac{x+3-1}{x+3}=\frac{6}{19}\)

\(19.\left(x+2\right)=6\left(x+3\right)\)

19x+38=6x+18

13x= -20

x= \(\frac{-20}{13}\)

1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

=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

=1/2-1/10

=5/10-1/10

=4/10=2/5

\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{8x9}+\frac{1}{9x10}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(\frac{1}{2}-\frac{1}{10}\)

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

= 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

= 1/2-0-0-0-0-0-0-0-0-1/10

=1/2 - 1/10 

=2/5

tớ trước nhé

a) \(5\frac{8}{17}:x+\frac{-1}{17}:x+3\frac{1}{17}:17\frac{1}{3}=\frac{4}{17}\)

\(\frac{93}{17}:x+\frac{-1}{17}:x+\frac{52}{17}:\frac{52}{3}=\frac{4}{17}\)

\(\left(\frac{93}{17}+\frac{-1}{17}\right):x+\frac{52}{17}.\frac{3}{52}=\frac{4}{17}\)

\(\frac{92}{17}:x+\frac{3}{17}=\frac{4}{17}\)

\(\frac{92}{17}:x=\frac{4}{17}-\frac{3}{17}\)

\(\frac{92}{17}:x=\frac{1}{17}\)

\(x=\frac{92}{17}:\frac{1}{17}\)

\(x=92\)

b) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{4}\right)+\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{1}{3}.\left(\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(1-\frac{1}{x+3}=\frac{6}{19}:\frac{1}{3}\)

\(1-\frac{1}{x+3}=\frac{18}{19}\)

\(\frac{1}{x+3}=1-\frac{18}{19}\)

\(\frac{1}{x+3}=\frac{1}{19}\)

\(\Rightarrow x+3=19\)

\(\Rightarrow x=19-3\)

\(\Rightarrow x=16\)

Bài 1 :

\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(S=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)

Bài 2 :

\(S=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)

\(S=\frac{1}{10}-\frac{1}{61}=\frac{51}{610}\)

Bài 3 :

\(3S=\frac{3}{4\times7}+\frac{3}{7\times11}+...+\frac{3}{19\times22}\)

\(3S=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{22}\)

\(3S=\frac{1}{4}-\frac{1}{22}\)

\(S=\frac{18}{88}\div3=\frac{6}{88}\)

A và B dễ 

Bài 2:

sai đề bài vì ngay từ cái phép tính đầu đã ko theo quy luật rồi 

\(A=\frac{-3}{5}-\frac{2}{5}+2\)

\(A=-1+2=1\)

\(B=\left(6-\frac{14}{5}\right).\frac{25}{8}-\frac{8}{5}=\frac{1}{4}\)

nÀ NÍ sao lại = đây là dấu trừ hay cộng 1/4

;1/4 nha