\(\left(x-\frac{1}{2004}\right)+\left(x-\frac{2}{2003}\right)-\left(x-\frac{3}{2002}\right)=x-\frac{4}{2001}\)

\(x-\frac{1}{2004}+x-\frac{2}{2003}-x+\frac{3}{2002}-x=-\frac{4}{2001}\)

\(x+x-x-x-\frac{1}{2004}-\frac{2}{2003}+\frac{3}{2002}=-\frac{4}{2001}\)

\(0x-\frac{1}{2004}-\frac{2}{2003}+\frac{3}{2002}=-\frac{4}{2001}\)

\(\Rightarrow\) Vô lý

Vậy \(x\in\phi\)

đúng ko zay

ko hieu de x-1 hay la xhi x-

trừ 2 vào mỗi tỉ số

xin loi minh moi hoc lop 6 thoi

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

\(\frac{x-1}{2004}+\frac{x-2}{2003}=\frac{x-3}{2002}+\frac{x-4}{2001}\)

\(\Rightarrow\frac{x-1}{2004}-1+\frac{x-2}{2003}-1=\frac{x-3}{2002}-1+\frac{x-4}{2001}-1\)

\(\Rightarrow\frac{x-2005}{2004}+\frac{x-2005}{2003}=\frac{x-2005}{2002}+\frac{x-2005}{2001}\)

\(\Rightarrow\frac{x-2005}{2001}+\frac{x-2005}{2002}-\frac{x-2005}{2003}-\frac{x-2005}{2004}=0\)

\(\Rightarrow\left(x-2005\right).\left(\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)

Vì \(\frac{1}{2001}>\frac{1}{2003};\frac{1}{2002}>\frac{1}{2004}\)

\(\Rightarrow\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\ne0\)

\(\Rightarrow x-2005=0\)

\(\Rightarrow x=2005\)

x+2005=0

=> ( x-2001/2 + ) + ( x-2002/3 + 1 ) = ( x-2003/4 + 1 ) + ( x-2004/5 + 1 )

=> x-1999/2 + x-1999/3 = x-1999/4 + x-1999/5

=> x-1999/2 + x-1999/3 - x-1999/4 - x-1999/5 = 0

=> (x-1999).(1/2+1/3-1/4-1/5) = 0

=> x-1999=0 ( vì 1/2+1/3-1/4-1/5 > 0 )

=> x = 1999

Vậy x = 1999

Tk mk nha

\(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\)

=> \(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}-\frac{x-4}{2001}=0\)

=> \(\left(\frac{x-1}{2004}-1\right)+\left(\frac{x-2}{2003}-1\right)-\left(\frac{x-3}{2002}-1\right)-\left(\frac{x-4}{2001}-1\right)=0\)

=> \(\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)

=> \(\left(x-2005\right).\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)

Vì \(\frac{1}{2004}< \frac{1}{2002}\); \(\frac{1}{2003}< \frac{1}{2001}\)

=> \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\)

=> \(x-2005=0\)

=> \(x=2005\)

Vậy \(x=2005\)