tim x:
(x-1)2016+(x-2)2016=1
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T7
2
PN
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
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2018}\)
\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1008}{1009}\)
\(2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1008}{1009}\)
\(2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1008}{1009}\)
\(2.\left(\dfrac{1}{2}-\dfrac{1}{x-1}\right)\) = \(\dfrac{1008}{1009}\)
\(\dfrac{1}{2}-\dfrac{1}{x-1}=\dfrac{504}{1009}\)
\(\dfrac{1}{x-1}=\dfrac{1}{2018}\)
\(x-1=2018\)
\(x=2019\)
PA
0
LD
7 tháng 5 2016
VẾ TRÁI LÀ:
A=1/2.3+1/3.2+1/4.5+...+1/x[x+1]
A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/n-1/n+1
A=1-1/n+1
1-1/n+1=2015/2016
1/n+1=1-2015/2016
1/n+1=1/2016
n=2016-1
n=2015
29 tháng 5 2016
Vế trái là
S=1/2+1/6+1/12+1/20+..+1/x(x+1)
S=1/1.2+1/2.3+1/3.4+1/4.5+...+1/x.(x+1)
S=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/x.(x+1)
S=1-1/(x+1)=Vế phải=2015/2016
<=>1-1/(x+1)=2015/2016
1/(x+1)=1/2016
=>x+1=2016
x=2015
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