tim x :1-1/x-1/3=1/6
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VN
1
NH
1
NH
4
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
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2018}\)
\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1008}{1009}\)
\(2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1008}{1009}\)
\(2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1008}{1009}\)
\(2.\left(\dfrac{1}{2}-\dfrac{1}{x-1}\right)\) = \(\dfrac{1008}{1009}\)
\(\dfrac{1}{2}-\dfrac{1}{x-1}=\dfrac{504}{1009}\)
\(\dfrac{1}{x-1}=\dfrac{1}{2018}\)
\(x-1=2018\)
\(x=2019\)
H
1
DP
17 tháng 7 2017
\(6\left(x+1\right)^2-2\left(x+1\right)^3-2\left(x-1\right)\left(x^2+x+1\right)=1\)
\(\Leftrightarrow6\left(x^2+2x+1\right)-2\left(x^3+3x^2+3x+1\right)-2\left(x^3+x^2+x-x^2-x-1\right)=1\)
\(\Leftrightarrow6x^2+12x+6-2x^3-6x^2-6x-2-2x^3-2x^2-2x+2x^2+2x+2=1\)
\(\Leftrightarrow-4x^3+6x+5=0\)
\(\Leftrightarrow x=1.5233401602\)
DT
0
TT
0
ND
0