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28 tháng 8 2016

x = 17 

còn giải thì dài dòng lắm 

28 tháng 8 2016

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

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

2 × [1/6×7 + 1/7×8 + 1/8×9 + ... + 1/x(x + 1)] = 2/9

1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + ... + 1/x - 1/x + 1 = 2/9 : 2

1/6 - 1/x + 1 = 2/9 × 1/2 = 1/9

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

1/x + 1 = 1/18

=> x + 1 = 18

=> x = 18 - 1 = 17

15 tháng 3 2017

Ta có:

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

\(\Leftrightarrow A=\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.9}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow A=\dfrac{1.2}{2.3.7}+\dfrac{1.2}{2.4.7}+\dfrac{1.2}{2.4.9}+...+\dfrac{1.2}{x\left(x+1\right)}=\dfrac{2}{9}\)

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

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

\(\Leftrightarrow A=2\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)

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

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

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

\(\Leftrightarrow x+1=18\)

\(\Leftrightarrow x=17\)

Vậy \(x=17\)

9 tháng 8 2017

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

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

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

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

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

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

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

\(\frac{1}{x+1}=\frac{1}{18}\Rightarrow18.1=1\left(x+1\right)\)

\(\Rightarrow18=x+1\Rightarrow x=18-1=17\)

13 tháng 8 2017

Cảm ơn bạn nha!

27 tháng 9 2020

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

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

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

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

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

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

=> \(\frac{1}{x+1}=\frac{1}{18}\)

=> x + 1 = 18

=> x = 17

Vậy x = 17

18 tháng 9 2020

\(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+.......+\frac{2}{x\left(x+1\right)}=\frac{11}{40}\)

\(\Leftrightarrow\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+........+\frac{2}{x\left(x+1\right)}=\frac{11}{40}\)

\(\Leftrightarrow2.\left[\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+.......+\frac{1}{x\left(x+1\right)}\right]=\frac{11}{40}\)

\(\Leftrightarrow\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+......+\frac{1}{x\left(x+1\right)}=\frac{11}{80}\)

\(\Leftrightarrow\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+.......+\frac{1}{x\left(x+1\right)}=\frac{11}{80}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+.......+\frac{1}{x}-\frac{1}{x+1}=\frac{11}{80}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{11}{80}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{16}\)

\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\)

Vậy \(x=15\)

16 tháng 7 2023

\(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\) +...+ \(\dfrac{2}{x\left(x+1\right)}\) = \(\dfrac{11}{40}\) (\(x\in\) N*)

\(\dfrac{1}{2}\).(\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+\(\dfrac{1}{28}\)+\(\dfrac{1}{36}\)+.....+ \(\dfrac{2}{x\left(x+1\right)}\)) = \(\dfrac{11}{40}\) \(\times\) \(\dfrac{1}{2}\)

\(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)

\(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)

\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+...+ \(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)

\(\dfrac{1}{5}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)

         \(\dfrac{1}{x+1}\) = \(\dfrac{1}{5}\) - \(\dfrac{11}{80}\)

           \(\dfrac{1}{x+1}\) = \(\dfrac{1}{16}\)

            \(x\) + 1 = 16

            \(x\)       = 16 - 1

             \(x\)     = 15 

27 tháng 9

Bài 1:

a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)

     \(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)

     \(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)

     \(x=-\) \(\dfrac{49}{56}\)

Vậy \(x=-\dfrac{49}{56}\)

b; 6 - \(x\) = - \(\dfrac{3}{4}\)

         \(x\) = 6 + \(\dfrac{3}{4}\)

         \(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)

         \(x=\dfrac{27}{4}\)

Vậy \(x=\dfrac{27}{4}\) 

c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)

              \(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)

              \(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)

               \(x=\dfrac{19}{20}\)

Vậy \(x=\dfrac{19}{20}\) 

27 tháng 9

      Bài 1:

d; - 6 - \(x\) = - \(\dfrac{3}{5}\)

      \(x\)   = - 6 + \(\dfrac{3}{5}\)

       \(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)

       \(x=-\dfrac{27}{5}\)

Vậy \(x=-\dfrac{27}{5}\)

e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)

             \(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)

             \(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)

              \(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)

               \(x=\dfrac{22}{21}\)

Vậy \(x=\dfrac{22}{21}\) 

f; - 8 - \(x\) =  - \(\dfrac{5}{3}\)

          \(x\) = \(-\dfrac{5}{3}\) + 8

         \(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)

         \(x\) = \(\dfrac{-19}{3}\)

Vậy \(x=-\dfrac{19}{3}\)