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10 tháng 1 2016
\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
\(\Leftrightarrow\frac{2-x}{2001}+1=\frac{1-x}{2002}+1+\frac{-x}{2003}+1\)
\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
\(\Leftrightarrow\frac{2003-x}{2001}-\frac{2003-x}{2002}-\frac{2003-x}{2003}=0\)
\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow2003-x=0\left(\text{ vì }\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\right)\)
<=>x=2003
Vậy S={2003}
KT
1
30 tháng 3 2017
\(\dfrac{x-4}{2001}\)- 1 +\(\dfrac{x-3}{2002}\)-1 + \(\dfrac{x-2}{2003}\)-1 =\(\dfrac{x-2003}{2}\)-1 + \(\dfrac{x-2002}{3}\)-1 +\(\dfrac{x-2001}{4}\)-1 <=> \(\dfrac{x-2005}{2001}\)+\(\dfrac{x-2005}{2002}\)+\(\dfrac{x-2005}{2003}\)-\(\dfrac{x-2005}{2}\)-\(\dfrac{x-2005}{3}\)-\(\dfrac{x-2005}{4}\)= 0 <=> (x-2005). (\(\dfrac{1}{2001}\)+\(\dfrac{1}{2002}\)+\(\dfrac{1}{2003}\)-\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)) =0 <=> x-2005=0 ( vì \(\dfrac{1}{2001}\) +\(\dfrac{1}{2002}\) +\(\dfrac{1}{2003}\)- \(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)- \(\dfrac{1}{4}\) khác 0) =>x = 2005
30 tháng 3 2017
x-4/2001+ x-3/2002 + x-2/2003= x-2003/2 + x-2002/3 + x-2001/4
<=>(x-4/2001 -1)+(x-3/2002 -1)+(x-2/2003 -1)-(x-2003/2 -1)+
(x-2002/3 -1)+(x-2001/4 -1) =0
<=>x-2005/2001+ x-2005/2002+ x-2005/2003- x-2005/2-
x-2005/3- x-2005/4 =0
<=>(x-2005).(1/2001+1/2002+1/2003- 1/2- 1/3- 1/4)=0
<=>x-2005=0 (vì 1/2001+1/2002+1/2003-1/2-1/3-1/4)
<=>x=2005
Vậy pt có nghiệm là x=2005
10 tháng 8 2016
\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
\(\Leftrightarrow\frac{2-x}{2001}+1=\left(\frac{1-x}{2002}+1\right)+\left(\frac{-x}{2003}+1\right)\)
\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow\) \(x=2003\)
10 tháng 8 2016
↔ \(\frac{2-x}{2001}+1\)\(=\left(\frac{1-x}{2002}+1\right)+\left(\frac{x}{2003}+1\right)\)
↔ \(\frac{2003-x}{2001}\) \(=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
↔ \(\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
↔ x = 2003
\(\dfrac{x-2}{2001}+\dfrac{x}{2003}=1+\dfrac{1-x}{2002}\Leftrightarrow\dfrac{x-2}{2001}+\dfrac{x}{2003}-\dfrac{x-1}{2002}-1=0\)
\(\Leftrightarrow\dfrac{x-2}{2001}-1+\dfrac{x}{2003}-1-\dfrac{x-1}{2002}+1=0\)
\(\Leftrightarrow\dfrac{x-2003}{2001}+\dfrac{x-2003}{2003}-\left(\dfrac{x-2003}{2002}\right)=0\)
\(\Leftrightarrow\left(x-2003\right)\left(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}\right)=0\) \(\Leftrightarrow x=2003\) vì \(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}>0\)Vậy...
Ta có: \(\dfrac{x-2}{2001}+\dfrac{x}{2003}=1+\dfrac{1-x}{2002}\)
\(\Leftrightarrow\dfrac{x-2}{2001}+\dfrac{x}{2003}-1+\dfrac{1-x}{2002}=0\)
\(\Leftrightarrow\dfrac{x-2}{2001}-1+\dfrac{x}{2003}-1+\dfrac{1-x}{2002}+1=0\)
\(\Leftrightarrow\dfrac{x-2003}{2001}+\dfrac{x-2003}{2003}-\dfrac{x-2003}{2002}=0\)
\(\Leftrightarrow\left(x-2003\right)\left(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}\right)=0\)
mà \(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}\ne0\)
nên x-2003=0
hay x=2003
Vậy: S={2003}