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DP
3 tháng 8 2017
\(\frac{x+4}{2001}+\frac{x+3}{2002}=\frac{x+2}{2003}+\frac{x+1}{2004}\)
\(\Leftrightarrow\left(\frac{x+4}{2001}+1\right)+\left(\frac{x+3}{2002}+1\right)=\left(\frac{x+2}{2003}+1\right)+\left(\frac{x+1}{2004}+1\right)\)
\(\Leftrightarrow\frac{x+2005}{2001}+\frac{x+2005}{2002}=\frac{x+2005}{2003}+\frac{x+2005}{2004}\)
\(\Leftrightarrow\frac{x+2005}{2001}+\frac{x+2005}{2002}-\frac{x+2005}{2003}-\frac{x+2005}{2004}=0\)
\(\Leftrightarrow\left(x+2005\right).\left(\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+\frac{1}{2004}\right)=0\)
Vì \(\left(\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+\frac{1}{2004}\right)\ne0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=0-2004=-2004\)
QQ
5
1 tháng 8 2015
\(\left(x-\frac{1}{2004}\right)+\left(x-\frac{2}{2003}\right)-\left(x-\frac{3}{2002}\right)=x-\frac{4}{2001}\)
\(x-\frac{1}{2004}+x-\frac{2}{2003}-x+\frac{3}{2002}-x=-\frac{4}{2001}\)
\(x+x-x-x-\frac{1}{2004}-\frac{2}{2003}+\frac{3}{2002}=-\frac{4}{2001}\)
\(0x-\frac{1}{2004}-\frac{2}{2003}+\frac{3}{2002}=-\frac{4}{2001}\)
\(\Rightarrow\) Vô lý
Vậy \(x\in\phi\)
V
1
V
1
QQ
0
HV
0
QQ
6
14 tháng 1 2016
x+4/2000+1+x+3/2001+1=x+2/2002+1+x+1/2... y cho nay la cong voi 1/1 chu lkhong phai la cong 1 o duoi mau dau nhe) quy dong chuyen ve ta duoc:
x+2004/2000+x+2004/2001-x+2004/2002-x+...
(x+2004)(1/2000+1/2001-1/2002-1/2003)=...
do 1/2000+1/2001-1/2002-1/2003 luon lon hon 0 nen suy ra:
x+2004=0 suy ra x=-2004
14 tháng 1 2016
x+4/2000+1+x+3/2001+1=x+2/2002+1+x+1/2... y cho nay la cong voi 1/1 chu lkhong phai la cong 1 o duoi mau dau nhe) quy dong chuyen ve ta duoc:
x+2004/2000+x+2004/2001-x+2004/2002-x+...
(x+2004)(1/2000+1/2001-1/2002-1/2003)=...
do 1/2000+1/2001-1/2002-1/2003 luon lon hon 0 nen suy ra:
x+2004=0 suy ra x=-2004
LM
0
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2004}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2004}{2005}\)
=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2004}{2005}\)
=> \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2004}{2005}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2004}{2005}:2=\frac{1002}{2005}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{1002}{2005}=\frac{1}{4010}\)
=> \(x+1=4010\)
=> \(x=4010-1\)
=> \(x=4009\)