\(c,\frac{x-a-b}{c}-1+\frac{x-b-c}{a}-1+\frac{x-a-c}{b}-1=0.\)

\(\frac{x-a-b-c}{c}+\frac{x-a-b-c}{a}+\frac{x-a-b-c}{b}=0\)

\(\left(x-a-b-c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)

=>\(\orbr{\begin{cases}a+b+c=x\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\end{cases}}\)

Vậy.......

[(2-x)/2001] -1 = [(1-x)/2002]-1 - [x/2003]+1

(2003-x) /2001 = (2003-x)/2002 - (2003-x)/2003

(2003-x)(1/2001-1/2002+1/2003)=0

x= 2003

mk chac chan 100% lun do

a,\(\Leftrightarrow\left(\frac{1-x}{2013}+1\right)=\left(\frac{2-x}{2012}+1\right)-\left(1-\frac{x}{2014}\right)\)

   \(\Leftrightarrow\frac{2014-x}{2013}=\frac{2014-x}{2012}-\frac{2014-x}{2014}\)

   \(\Leftrightarrow\frac{2014-x}{2013}-\frac{2014-x}{2012}+\frac{2014-x}{2014}\)=0

   \(\Leftrightarrow\left(2014-x\right)\left(\frac{1}{2013}-\frac{1}{2012}+\frac{1}{2014}\right)=0\)

   \(\Leftrightarrow x=2014\left(do.cái.còn.lại.\ne0\right)\)

b,tương tự +1 vào cái thứ nhất ,+1 vào cái thứ 2,1- vào cái thứ 3 được x=2013

ban oi them bot sai roi

\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)

\(\Leftrightarrow\frac{2-x}{2001}+1=\left(\frac{1-x}{2001}+1\right)+\left(\frac{-x}{2003}+1\right)\)

\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)

\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow\left(2003-x\right)=0\) (vì \(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\))

\(\Leftrightarrow x=2003\).

Vậy tập nghiệm của phương trình là \(S=\left\{2003\right\}\).

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

\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2004}-\frac{x+2005}{2003}-\frac{x+2005}{2003}=0\)

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

\(\Leftrightarrow x+2005=0\Leftrightarrow x=-2005\) 

=> (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)(1/2004+1/2003-1/2002-1/2001)=0
=> x+2005=0
=> x=-2005

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

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

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

\(\Leftrightarrow x+2005=0\)

\(\Leftrightarrow x=-2005\)

b) Sửa đề :

\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)

\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)

\(\Leftrightarrow x=300\)

c) \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)

\(\Leftrightarrow\frac{2-x}{2002}+1=\frac{1-x}{2003}+1-\frac{x}{2004}+1\)

\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}\)

\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)=0\)

\(\Leftrightarrow x=2004\)

Vậy....

a, Ta có : \(\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)

=> \(\frac{4\left(x+1\right)}{12}+\frac{9\left(2x+1\right)}{12}=\frac{2\left(2x+3\left(x+1\right)\right)}{12}+\frac{7+12x}{12}\)

=> \(4\left(x+1\right)+9\left(2x+1\right)=2\left(2x+3\left(x+1\right)\right)+7+12x\)

=> \(4\left(x+1\right)+9\left(2x+1\right)=2\left(2x+3x+3\right)+7+12x\)

=> \(4x+4+18x+9=4x+6x+6+7+12x\)

=> \(4x+18x-12x-6x-4x=6+7-4-9\)

=> \(0x=0\) ( Luôn đúng với mọi x )

Vậy phương trình có vô số nghiệm .

b, Ta có : \(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)

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

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

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

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

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

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

=> \(2003-x=0\)

=> \(x=2003\)

Vậy phương trình có tập nghiệm là \(S=\left\{2003\right\}\)

\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)

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

\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)

\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow\) \(x=2003\) 

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

↔ \(\frac{2003-x}{2001}\) \(=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)

↔ \(\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

↔ x = 2003