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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
NP
0
MH
2
9 tháng 8 2017
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{2}{9}.\frac{1}{2}\)
\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
\(\frac{1}{x+1}=\frac{1}{18}\Rightarrow18.1=1\left(x+1\right)\)
\(\Rightarrow18=x+1\Rightarrow x=18-1=17\)
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 \(x=17\)
XO
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
NN
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 \(x=15\)
