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30 tháng 3 2017
\(\frac{1}{3}\) + \(\frac{1}{6}\) + \(\frac{1}{10}\) + ... + \(\frac{1}{x\left(x+1\right):2}\)
= \(\left(1-\frac{1}{2018}\right)-\frac{1}{2018}\)
= \(\frac{2017}{2018}-\frac{1}{2018}\)
= \(\frac{2016}{2018}=\frac{1008}{1009}\)
10 tháng 5 2017
1/2.(1/3+1/6+1/10+...+1/x(x+1))=1/2.2016/2018
1/6+1/12+1/20+...+1/x(x+1)=504/1009
1/2.3+1/3.4+1/4.5+...+1/x(x+1)=504/1009
1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=504/1009
1/2-1/x+1=504/1009
x-1/2(x+1)=504/1009
-> 1009(x-1)=504.2(x+1)
1009x-1009=1008x+1008
1009x-1008x=1008+1009
->x=2017
LB
10 tháng 5 2017
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2016}{2018}\)
\(A=\frac{1}{2\left(2+1\right):2}+\frac{1}{3\left(3+1\right):2}+...+\frac{1}{x\left(x+1\right):2}\)
\(A=\frac{1}{2\left(2+1\right)}\cdot2+\frac{1}{3\left(3+1\right)}\cdot2+...+\frac{1}{x\left(x+1\right)}.2=\frac{2016}{2018}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2016}{2018}\)
\(A=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2016}{2018}\)
\(A=1-\frac{1}{x+1}=\frac{2016}{2018}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2016}{2018}=\frac{1}{1009}\)
\(\Rightarrow x+1=1009\Rightarrow x=1008\)
MV
11 tháng 5 2017
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\cdot\left(x+1\right):2}=\dfrac{2016}{2018}\\ \dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\cdot\left(x+1\right)}=\dfrac{2016}{2018}\\ 2\cdot\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\cdot\left(x+1\right)}\right)=\dfrac{2016}{2018}\\ 2\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{x\cdot\left(x+1\right)}\right)=\dfrac{2016}{2018}\\ 2\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}:2\\ \dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1008}{2018}\\ \dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{1008}{2018}\\ \dfrac{1}{x+1}=\dfrac{1}{2018}\\ \Leftrightarrow x+1=2018\\ x=2018-1\\ x=2017\)
7 tháng 8 2018
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
NM
3
9 tháng 5 2017
Tìm số tự nhiên x biết
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\cdot\left(x+1\right):2}=\frac{2016}{2018}\)
N
8
D
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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