\(\frac{x-1009}{1001}\)+\(\frac{x-4}{1003}\)+\(\frac{x+2010}{1005}\)=7

⇔\(\frac{x-1009}{1001}\)+\(\frac{x-4}{1003}\)+\(\frac{x+2010}{1005}\)-7=0

⇔\(\left(\frac{x-1009}{1001}-1\right)+\left(\frac{x-4}{1003}-2\right)+\left(\frac{x+2010}{1005}-4\right)=0\)

⇔\(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)

⇔(x-2010)\(\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)\)=0

⇔x-2010=0

⇔x=2010

Vậy x=2010

\(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)

⇔ \(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}-7=0\)

⇔\(\left(\frac{x-1009}{1001}-1\right)+\left(\frac{x-4}{1003}-2\right)\)\(+\left(\frac{x+2010}{1005}-4\right)=0\)

⇔\(\frac{x-1009-1001}{1001}+\frac{x-4-2006}{1003}+\)\(\frac{x+2010-4020}{1005}=0\)

⇔\(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)

⇔\(\left(x-2010\right)\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)=0\)

⇔ \(x-2010=0\left(do\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}>0\right)\)

⇔ \(x=2010\)

Vậy S = {2010}

ta có : 

\(\frac{x-1009}{1001}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)

hay \(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\Leftrightarrow x-2010=0\)

hay x =2010

Vậy phương trình có nghiệm x = 2010

\(\Leftrightarrow\frac{x-1009}{1001}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)

\(\Leftrightarrow\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)=0\)

\(\Leftrightarrow x=2010\)

\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)

\(\Leftrightarrow\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}-7=0\)

\(\Leftrightarrow\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)

\(\Leftrightarrow\dfrac{x-2010}{1001}+\dfrac{x-2010}{1003}+\dfrac{x-2010}{1005}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)

\(\Leftrightarrow x-2010=0\)

\(\Rightarrow x=2010\)

Vậy....

\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)

\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}-7=0\)

\(\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)

\(\dfrac{x-2010}{1001}+\dfrac{x-2010}{1003}+\dfrac{x-2010}{1005}=0\)

\(\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)

\(x-2010=0\)

\(x=2010\)

Vậy x = 2010

\(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)

\(\Leftrightarrow\frac{x-1009}{1001}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)

\(\Leftrightarrow\frac{x-1009-1001}{1001}+\frac{x-4-2006}{1003}+\frac{x+2010-4020}{1005}=0\)

\(\Leftrightarrow\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)=0\)

\(\Leftrightarrow x-2010=0\)

\(\Leftrightarrow x=2010\)

V...\(S=\left\{2010\right\}\)

^^

\(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)

\(\Leftrightarrow\left(\frac{x-1009}{1001}-1\right)+\left(\frac{x-4}{1003}-2\right)+\left(\frac{x+2010}{1005}-4\right)=0\)

\(\Leftrightarrow\frac{x-1009-1001}{1001}+\frac{x-4-2006}{1003}+\frac{x+2010-4020}{1005}=0\)

\(\Leftrightarrow\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)=0\)

\(\Leftrightarrow x-2010=0\)

\(\Leftrightarrow x=2010\)

a)\(\frac{x-10}{2010}\)+ \(\frac{x-3}{2003}\)+\(\frac{x-2}{2002}\)= -3

=> \(\frac{x-10}{2010}\)+1+ \(\frac{x-3}{2003}\)+ 1+\(\frac{x-2}{2002}\)+1= -3 +1 + 1 + 1

=> \(\frac{x-10+2010}{2010}\)+ \(\frac{x-3+2003}{2003}\)+\(\frac{x-2+2002}{2002}\)= 0

=>\(\frac{x+2000}{2010}\)+ \(\frac{x+2000}{2003}\)+\(\frac{x+2000}{2002}\)= 0

=>(x + 2000)(\(\frac{1}{2010}\)+ \(\frac{1}{2003}\)+\(\frac{1}{2002}\)) = 0

=> x + 2000 = 0

hoặc

=>\(\frac{1}{2010}\)+ \(\frac{1}{2003}\)+\(\frac{1}{2002}\)= 0

Mà : \(\frac{1}{2010}\)> 0

\(\frac{1}{2003}\)> 0

\(\frac{1}{2002}\)> 0

Cộng vế theo vế của các bất đẳng thức trên , ta có:

\(\frac{1}{2010}\)+\(\frac{1}{2003}\)+\(\frac{1}{2002}\)>0

=> x + 2000 = 0

=> x = 0 -2000 = -2000

Vậy x = -2000

Nhường các bạn câu 2 :(

x-1009/1001+x-4/1003+x+2010/1005=7

((x-1009/1001)-1))+((x-4/1003)-2)+((x+2010/1005)-4))=0

(x-2010/1001)+(x-2010/1003)+(x-2010/1005)=0

(x-2010)*(1/1001+1/1003+1/1005)=0

okk!!!!!!!!!!!!!!!

Thanks bingodeo nhé :))