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19 tháng 4 2015

1.

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.............+\frac{1}{99.101}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{99}-\frac{1}{101}\)

= 1 - \(\frac{1}{101}\)

\(\frac{100}{101}\)

23 tháng 4 2016

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

23 tháng 4 2016

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

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

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

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

29 tháng 5 2017

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

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

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

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

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

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

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

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

\(\Rightarrow x+3=19\)

\(\Rightarrow x=19-3\)

\(\Rightarrow x=16\)

8 tháng 8 2017

\(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}\cdot\frac{17}{4}-28\cdot\frac{4}{3}\right):\frac{7}{4}\)

\(=\frac{59}{15}-\frac{29}{4}:\frac{7}{4}=\)\(\frac{59}{15}-\frac{29}{7}=\frac{-22}{105}\)

8 tháng 8 2017

B = \(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}x\frac{17}{4}-2x\frac{4}{3}\right):\frac{7}{4}\)

    = \(\frac{59}{10}x\frac{2}{3}-\left(\frac{119}{12}-\frac{8}{3}\right)x\frac{4}{7}\)

    = \(\frac{59}{15}-\frac{29}{4}x\frac{4}{7}=\frac{59}{15}-\frac{29}{7}\)

     = \(\frac{-22}{105}\)

C = \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\)

    = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)

     = \(1-\frac{1}{7}=\frac{6}{7}\)

11 tháng 4 2019

A và B dễ 

Bài 2:

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

11 tháng 4 2019

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

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

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

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

5 tháng 5 2016

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

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

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

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

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

19x+38=6x+18

13x= -20

x= \(\frac{-20}{13}\)

12 tháng 5 2020

Ta có :

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

12 tháng 5 2020

cảm ơn bạn nha

8 tháng 8 2016

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{96}\)

\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{96}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{5}{16}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{5}{16}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{5}{16}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{5}{16}\)

\(\frac{1}{2x+3}=\frac{1}{48}\)

=> 2x + 3 = 48

=> 2x = 48 - 3

=> 2x = 45

=> x = 45/2