NT
0
K
Khách
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LT
9 tháng 3 2020
\(\frac{1}{2013}.x+1+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{2012.2013}=2\)
\(\Rightarrow\frac{1}{2013}.x+1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}=2\)
\(\Rightarrow\frac{1}{2013}.x+1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013}.x+1+\frac{1}{1}-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013}.x+2-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013}.x=\frac{1}{2013}\Rightarrow x=1\)
Vậy x=1
CHÚC CÁC EM HỌC TỐT
EC
0
NN
2
16 tháng 7 2016
1/2013.x+1+1/2+1/6+1/12+...+1/2012.2013=2
1/2013.x+1+1/1.2+1/2.3+1/3.4+...+1/2012.2013=2
1/2013.x+1+1-1/2+1/2-1/3+1/3-1/4+...+1/2012-1/2013=2
1/2013.x+2-1/2013=2
1/2013.x =2-2+1/2013
1/2013.x =1/2013
=>2013.x=2013
=> x=1
16 tháng 7 2016
\(\Rightarrow\frac{1}{2013.x}+1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2012}-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013.x}+2-\frac{1}{2013}=2\)
\(\Rightarrow\frac{1}{2013.x}=2-2+\frac{1}{2013}\)
\(\Rightarrow\frac{1}{2013.x}=\frac{1}{2013}\)
\(\Rightarrow2013.x=2013\)
\(\Rightarrow x=1\)
TN
1 tháng 5 2016
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2015}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{4030}\)
tự làm tiếp nhé mk ăn cơm đã
NM
0
NT
1
NT
2
17 tháng 9 2016
1/2x3/2+1/3x4/2+1/4x5/2+1/5x6/2+.......+2/Xx(X+1)=2011/2013
2/2x3+2/3x4+2/4x5+2/5x6+.....+2/Xx(X+1)=2011/2013
2x(1/2x3+1/3x4+1/4x5+1/5x6+....+1/Xx(x+1)=2011/2013
1/2x3+1/3x4+1/4x5+1/5x6+....+1/Xx(X+1)=2011/4026
1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+.....+ 1/x-1/x+1=2011/4026
1/2-1/x+1=2011/4026
1/x+1=1/2-2011/4026
1/x+1=1/2013
Suy ra x=2012
SN
17
2 tháng 4 2015
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\cdot\left(x+1\right)}=\frac{2013}{2015}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2013}{2015}:2\)
\(\Rightarrow-\frac{1}{x+1}=\frac{2013}{4030}-\frac{1}{2}\)
\(\Rightarrow-\frac{1}{x+1}=-\frac{1}{2015}\Rightarrow x+1=2015\Rightarrow x=2014\)
TP
0
