Tìm x thuộc N
1\3+1\6+1\10+.....+2\x.(x+1)=1999\2001
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AD
2
HT
9 tháng 8 2015
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{2001}:2=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{2001}=\frac{1}{2001}\)
=> x + 1 = 2001
=> x = 2001 - 1
=> x = 2000
9 tháng 8 2015
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+..+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\frac{1}{6}+\frac{1}{12}+..+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}:\frac{1}{2}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\frac{1}{x+1}=\frac{1}{2001}\)
=> x + 1 = 2001
=> x = 2001 - 1
=> x = 2000
DC
3
9 tháng 8 2015
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+..+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}:2\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\frac{x+1-2}{2\left(x+1\right)}=\frac{1999}{4002}\Rightarrow\frac{x-1}{2\left(x+1\right)}=\frac{1999}{4002}\Leftrightarrow4002\left(x-1\right)=1999.2\left(x+1\right)\)
=> 4002x - 4002 = 3998x + 3998
=> 4002x - 3998x = 3998 + 4002
=> 4x = 8000
=> x = 2000
NT
24 tháng 3 2018
!/3+1/6+1/10+...+2/x(x+1)=1999/2001
1/6+1/12+1/20+...+2/x(x+1)=1999/2001
2(1/6+1/12+1/20+...+1/x(x+1)=1999/2001
1/6+1/12+1/20+1/x(x+1)=1999/2001:2
1/6+1/12+1/20+...+1/x(x+1)=1999/4002
1/2x3+1/3x4+1/4x5+...+1/x(x+1)=1999/4002
1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=1999/4002
1/2-1/x+1=1999/4002
1/x+1=1/2-1999/4002
1/x+1=1/2001
=>x+1=2001
=>x=2001-1
=x=2000
Vậy x=2000.
NK
2
19 tháng 12 2017
(*) <=> 1\6 + 1\12 +.. + 1\x.(x+1) = 2009\(2011.2)
ma
1\2.3 =1\2-1\3
1\3.4=1\3-1\4
...............
1\x(x+1)= 1\x-1\(x+1)
cong tung ve ta dc
Vt= 1\2- 1\(x+1) =2009\(2.2011)
<=> 2011\(2.2011) -2009\(2.2011) =1\(x+1)
<=> 1\2011 =1\(x+1)
=> x=2010
19 tháng 12 2017
1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 1999/2001
nhân 1/2 vào 2 vế ta được vế trái là :
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{x-1}{2.\left(x+1\right)}=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{x-1}{\left(x+1\right)}=\frac{1999}{2001}\)
suy ra : 2001x - 2001 = 1999x + 1999
2x = 1999 + 2001 = 4000
=> x = 2000
LL
1
13 tháng 5 2017
1/2x[1/3+1/6+1/10+....2/x.(x+1)]=1999/2001x1/2
1/2x3+1/3x4+....+1/x(x+1)=1999/4002
1/2-1/3+1/3-1/4+....+1/x-1/x+1=1999/4002
1/2-1/x+1=1999/4002
1/x+1=1/2-1999/4002
1/x+1=1/2001
=>x+1=2001
x=2001-1
x=2000
QD
3
PQ
28 tháng 8 2015
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1999}{2001}\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{2001}:2\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{2001}:2=\frac{1}{2001}\Rightarrow x+1=2001\Rightarrow x=2000\)
