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CTVHS
8 tháng 5

vế kia đâu

1/21 + 1/28 + 1/36 + .... + 2/x.(x+1) = ?

14 tháng 7 2015

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

2\(\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{9}:2=\frac{1}{9}\)

\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)

=> x+1 =18

=> x = 18 - 1

=> x = 17

4 tháng 8 2017

giúp với

4 tháng 8 2017

sorry nha ko biết làm

18 tháng 3 2016

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{6\cdot7}+\frac{2}{7\cdot8}+\frac{2}{8\cdot9}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}\div2\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)

\(\Rightarrow x+1=18\Rightarrow x=18-1\Rightarrow x=17\)

27 tháng 9 2016

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}\right)=\frac{1}{2}.\frac{2}{9}\)

\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x.\left(x+1\right)}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x.\left(x+1\right)}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)

\(\Rightarrow\frac{1}{x+1}=\frac{3}{18}-\frac{2}{18}=\frac{1}{18}\)

\(\Rightarrow x+1=18\)

\(\Rightarrow x=18-1=17\)

Vậy \(x=17\)

3 tháng 6 2017

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(2.\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x-1}\right)=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{x-1}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{x-1}=\frac{2}{9}:2\)

\(\frac{1}{6}-\frac{1}{x-1}=\frac{1}{9}\)

\(\frac{1}{x-1}=\frac{1}{6}-\frac{1}{9}\)

\(\frac{1}{x-1}=\frac{1}{18}\)

\(\Rightarrow x-1=18\)

\(\Rightarrow x=18+1\)

\(\Rightarrow x=19\)

3 tháng 6 2017

 = 2/42 + 2/56+2/72+................+2/x.(x+1)=2/9

=\(\frac{2}{6.7}\)+\(\frac{2}{7.8}\)+\(\frac{2}{8.9}\)+......+\(\frac{2}{x.\left(x+1\right)}\)=2/9

=2.( \(\frac{1}{6}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{8}\)+.......+\(\frac{1}{x}\)-\(\frac{1}{x+1}\)

=2.(1/6 -\(\frac{1}{x+1}\))=2/9

=1/6 -\(\frac{1}{x+1}\)=2/9:2=1/9

=1/6-1/9=\(\frac{1}{x+1}\)=3/54=1/18

=> x= 18-1 =17

27 tháng 5 2015

1/21 +1/28 +1/36 ...2/x (x+1)=2/9

=>1/3.7 +1/4.7+1/4.9 ....2/x.(x+1)=2/9

=>2/6.7+2/7.8+2/8.9....+2/x.(x+1)=2/9

=>2.(1/6-1/x+1)=2/9

=>1/6 - 1/x +1 =1/9

=>1/x + 1 =1/18

=>x+1=18

=>x=17 

DS:17 

8 tháng 4 2017

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NV
1 tháng 3 2022

\(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)

\(\Leftrightarrow x+1=18\)

\(\Leftrightarrow x=17\)