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\(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\)
=> \(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}-\frac{x-4}{2001}=0\)
=> \(\left(\frac{x-1}{2004}-1\right)+\left(\frac{x-2}{2003}-1\right)-\left(\frac{x-3}{2002}-1\right)-\left(\frac{x-4}{2001}-1\right)=0\)
=> \(\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)
=> \(\left(x-2005\right).\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
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 \(x=2005\)
Ta có: \(\frac{x-1}{2004}+\frac{x-2}{2003}=\frac{x-3}{2002}+\frac{x-4}{2001}\)
\(\Leftrightarrow\frac{x-1}{2004}-1+\frac{x-2}{2003}-1=\frac{x-3}{2002}-1+\frac{x-4}{2001}-1\)
\(\Leftrightarrow\frac{x-1-2004}{2004}+\frac{x-2-2003}{2003}=\frac{x-3-2002}{2002}+\frac{x-4-2001}{2001}\)
\(\Leftrightarrow\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)
\(\Leftrightarrow\left(x-2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Vì \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\)
=> x - 2005 = 0
=> x = 2005
Vậy x = 2005
=> (x - 1)/2004 - 1 + (x - 2)/2003 - 1 = (x - 3)/2002 -1 + (x - 4)/2001 - 1
=> (x - 2005)/2004 + (x - 2005)/2003 = (x - 2005)/2002 + (x - 2005)/2001
=> (x - 2005)/2004 + (x - 2005)/2003 - (x - 2005)/2002 - (x - 2005)/2001 = 0
=> (x - 2005) * ( 1/2004 + 1/2003 - 1/2002 - 1/2001) = 0
Ta thấy ( 1/2004 + 1/2003 - 1/2002 - 1/2001) khác 0
=> x - 2005 = 0
=> x = 2005
\(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\)
\(\Leftrightarrow\)\(\frac{x-1}{2004}+\frac{x-2}{2003}=\frac{x-4}{2001}+\frac{x-3}{2002}\)
\(\Leftrightarrow\)\(\frac{x-1}{2004}-1+\frac{x-2}{2003}-1=\)\(\frac{x-4}{2001}-1+\frac{x-3}{2002}-1\)
\(\Leftrightarrow\)\(\frac{x-2005}{2004}+\frac{x-2005}{2003}\)\(=\frac{x-2015}{2001}+\frac{x-2005}{2002}\)
\(\Leftrightarrow\)\(\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2001}-\frac{x-2005}{2002}=0\)
\(\Leftrightarrow\)( x - 2005 ) ( \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{2002}\)) = 0
Do \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{2002}\)\(\ne\)0
\(\Rightarrow\)x - 2005 = 0
\(\Leftrightarrow\)x = 2005
Vậy x = 2005
\(a.\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\left(\frac{x+4}{1997}+1\right)=0\)
\(=\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)
\(=\left(x+2001\right).\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)
\(=>x+2001=0\)
\(x=-2001\)
\(b.\left(\frac{x+1}{1999}-1\right)+\left(\frac{x+2}{2000}-1\right)+\left(\frac{x+3}{2001}-1\right)=\left(\frac{x+4}{2002}-1\right)+\left(\frac{x+5}{2003}-1\right)\)\(+\left(\frac{x+6}{2004}-1\right)\)
\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}=\frac{x+1998}{2002}+\frac{x+1998}{2003}+\frac{x+1998}{2004}\)
\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}-\frac{x+1998}{2002}-\frac{x+1998}{2003}-\frac{x+1998}{2004}=0\)
\(\left(x+1998\right).\left(\frac{1}{1999}+\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)
\(=>x+1998=0\)
\(x=-1998\)
dễ quá!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
\(\Leftrightarrow\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
De thay \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}< 0\Rightarrow x+2005=0\)
\(\Rightarrow x=-2005\)
Bài giải
\(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
\(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
\(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)
\(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Do : \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\)
\(\Rightarrow\text{ }x+2005=0\)
\(x=0-2005\)
\(x=-2005\)
\(\frac{x-1}{2005}+\frac{x-2}{2004}-\frac{x-3}{2003}=\frac{x-4}{2002}\)
=>\(\frac{x-1}{2005}+\frac{x-2}{2004}-\frac{x-3}{2003}-\frac{x-4}{2004}=0\)
=>\(\left(\frac{x-1}{2005}-1\right)+\left(\frac{x-2}{2004}-1\right)-\left(\frac{x-3}{2003}-1\right)-\left(\frac{x-4}{2002}-1\right)=0\)
=>\(\frac{x-1-2005}{2005}+\frac{x-2-2004}{2004}-\frac{x-3-2003}{2003}-\frac{x-4-2002}{2002}=0\)
=>\(\frac{x-2006}{2005}+\frac{x-2006}{2004}-\frac{x-2006}{2003}-\frac{x-2006}{2002}=0\)
=>\(\left(x-2006\right)\left(\frac{1}{2005}+\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
Mà \(\frac{1}{2005}+\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\ne0\)
=> x - 2006 = 0 => x = 2006
x=-2007