x-1/2004+x-2/2003-x-3/2002=x-4/2001
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KT
1
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 \(x\in\phi\)
4 tháng 10 2016
\(\frac{x-1}{2004}+\frac{x-2}{2003}=\frac{x-3}{2002}+\frac{x-4}{2001}\)
\(\Rightarrow\frac{x-1}{2004}-1+\frac{x-2}{2003}-1=\frac{x-3}{2002}-1+\frac{x-4}{2001}-1\)
\(\Rightarrow\frac{x-2005}{2004}+\frac{x-2005}{2003}=\frac{x-2005}{2002}+\frac{x-2005}{2001}\)
\(\Rightarrow\frac{x-2005}{2001}+\frac{x-2005}{2002}-\frac{x-2005}{2003}-\frac{x-2005}{2004}=0\)
\(\Rightarrow\left(x-2005\right).\left(\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)
Vì \(\frac{1}{2001}>\frac{1}{2003};\frac{1}{2002}>\frac{1}{2004}\)
\(\Rightarrow\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\ne0\)
\(\Rightarrow x-2005=0\)
\(\Rightarrow x=2005\)
V
1
V
1
QQ
0
KK
4
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 \(\left(\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+\frac{1}{2004}\right)\ne0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=0-2004=-2004\)
PQ
1
28 tháng 3 2020
- Ta có : \(\frac{x-1}{2004}+\frac{x-2}{2003}=\frac{x-3}{2002}+\frac{x-4}{2001}\)
=> \(\frac{x-1}{2004}-1+\frac{x-2}{2003}-1=\frac{x-3}{2002}-1+\frac{x-4}{2001}-1\)
=> \(\frac{x-2005}{2004}+\frac{x-2005}{2003}=\frac{x-2005}{2002}+\frac{x-2005}{2001}\)
=> \(\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)
=> \(\left(x-2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
=> \(x-2005=0\)
=> \(x=2005\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{2005\right\}\)
