1/3 +1/6+ 1/10 + .......+2/x(x+1 ) = 2016/2018
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KS
14 tháng 5 2022
số số hạng của mẫu dãu số :
`(2018 - 2) : 1 + 1= 2017(số hạng)`
tổng mãu dãu trên là :
` (2018 + 2) xx 2017 : 2 = 2037170`
=> `1/2 + 1/3 + 1/4 + ... + 1/2018 = 1/2037170`
14 tháng 5 2022
-Chỉ áp dụng bài này với số tự nhiên có quy luật thôi bạn nhé!
19 tháng 3 2022
\(A=\left(a\text{x}7+a\text{x}8-a\text{x}15\right):\left(1+2+3+...+10\right)\)
\(A=\left(a\text{x}\left(7+8-15\right)\right):\left(1+2+3+...+10\right)\)
\(A=\left(a\text{x}0\right):\left(1+2+3+..+10\right)\)
\(A=0:\left(1+2+3+...+10\right)\)
\(A=0\)
\(B=\left(18-9\text{x}2\right)\text{x}\left(2+4+6+8+10\right)\)
\(B=\left(18-18\right)\text{x}\left(2+4+6+8+10\right)\)
\(B=0\text{x}\left(2+4+6+8+10\right)\)
\(B=0\)
NT
2
1 tháng 1 2018
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2017}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2017}\)
\(\Leftrightarrow\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+..........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2018}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1008}{2018}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2018}\)
\(\Leftrightarrow x+1=2018\)
\(\Leftrightarrow x=2017\)
Vậy ..
1 tháng 1 2018
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2018}\)
\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1008}{1009}\)
\(2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1008}{1009}\)
\(2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1008}{1009}\)
\(2.\left(\dfrac{1}{2}-\dfrac{1}{x-1}\right)\) = \(\dfrac{1008}{1009}\)
\(\dfrac{1}{2}-\dfrac{1}{x-1}=\dfrac{504}{1009}\)
\(\dfrac{1}{x-1}=\dfrac{1}{2018}\)
\(x-1=2018\)
\(x=2019\)
30 tháng 3 2017
\(\frac{1}{3}\) + \(\frac{1}{6}\) + \(\frac{1}{10}\) + ... + \(\frac{1}{x\left(x+1\right):2}\)
= \(\left(1-\frac{1}{2018}\right)-\frac{1}{2018}\)
= \(\frac{2017}{2018}-\frac{1}{2018}\)
= \(\frac{2016}{2018}=\frac{1008}{1009}\)
27 tháng 12 2017
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2016}{2018}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{504}{1009}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{504}{1009}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{504}{1009}\)
\(\frac{1}{x+1}=\frac{1}{2018}\)
\(\Rightarrow x+1=2018\)
\(\Rightarrow x=2017\)
TT
0
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2016}{2018}\)
<=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1008}{1009}\)
<=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1008}{1009}\)
<=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{504}{1009}\)
<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{504}{1009}\)
<=> \(\frac{1}{x+1}=\frac{1}{2018}\)
=> \(x+1=2018\)
<=> \(x=2017\)
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