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
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
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
x+4/2000+x+3/2001=x+2/2002+x+1/2003
cong 2 vao hai ve ta duoc
x+4/2000+1+x+3/2001+1=x+2/2002+1+x+1/2...
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
x-1/2004+x-2/2003-x-3/2002=x-4/2001
{x-1/2004-1}+{x-2/2003}-{x-3/2002}-{x-4/2001}=0
{x-2005/2004}+{x-2005/2003}-{x-2005/2002}-{x-2005/2001}
x-2005(1/2004-1/2003+1/2002+1/2001)
vi (1/2004-1/2003+1/2002+1/2001) luon kha
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
=>\(\frac{x+4}{2000}+\frac{x+3}{2001}+2=\frac{x+2}{2002}+\frac{x+1}{2003}+2\)
=>\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
=>\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\)
mà \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}>0\)
=>\(x+2004=0\)
=>\(x=-2004\)