Trong S₁ có các số chia hết cho các thừa số ở S₂

< = > S₁ chia hết cho S₂

=> ĐPCM 

\(\dfrac{x-1001}{1006}+\dfrac{x-1003}{1004}+\dfrac{x-1005}{1002}+\dfrac{x-1007}{1000}=4\)

\(\Rightarrow\dfrac{x-1001}{1006}-1+\dfrac{x-1003}{1004}-1+\dfrac{x-1005}{1002}-1+\dfrac{x-1007}{1000}-1=0\)

\(\Rightarrow\dfrac{x-2007}{1006}+\dfrac{x-2007}{1004}+\dfrac{x-2007}{1002}+\dfrac{x-2007}{1000}=0\)

\(\Rightarrow\left(x-2007\right)\left(\dfrac{1}{1006}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}\right)=0\)

Dễ thấy: \(\dfrac{1}{1000}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}>0\Leftrightarrow x-2007=0\Leftrightarrow x=2007\)

\(\dfrac{x-1016}{1001}+\dfrac{x-13}{1002}+\dfrac{x+992}{1003}=\dfrac{x+995}{1004}+\dfrac{x-7}{1005}+1\)

<=>\(\dfrac{x-1016}{1001}-1+\dfrac{x-13}{1002}-2+\dfrac{x+992}{1003}-3=\dfrac{x+995}{1004}-3+\dfrac{x-7}{1005}-2\)

<=>\(\dfrac{x-2017}{1001}+\dfrac{x-2017}{1002}+\dfrac{x-2017}{1003}=\dfrac{x-2017}{1004}+\dfrac{x-2017}{1005}\)

<=>\(\left(x-2017\right)\left(\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}-\dfrac{1}{1004}-\dfrac{1}{1005}\right)=0\)

vì 1/1001+1/1002+1/1003-1/1004-1/1005 khác 0 nên x-2017=0<=>x=2017

vậy..........

x+2/5+x+2/6+x+2/7=x+2/8+x+2/9                                                   
x+107/105=17/36                                                                         
x=17/36-107/105                                                                             
x=-689/1260                                                                                  
 

câu kia mik ko thik làm

\(a.x+\frac{2}{5}+x+\frac{2}{6}+x+\frac{2}{7}=x+\frac{2}{8}+x+\frac{2}{9}\)

\(3x+\left(\frac{2}{5}+\frac{2}{6}+\frac{2}{7}\right)=2x+\left(\frac{2}{8}+\frac{2}{9}\right)\)

\(3x-2x=\left(\frac{2}{5}+\frac{2}{6}+\frac{2}{7}\right)-\left(\frac{2}{8}+\frac{2}{9}\right)\)

\(1x=\frac{689}{1260}=0,54\)