`a) S=1-2-3+4+5-6-7+8+...+2001-2002-2003+2004+2005`
`-> S=(1-2-3+4)+(5-6-7+8)+...+(2001-2002-2003+2004)+2005`
`-> S=0+0+...+0+2005`
`-> S=2005`
`b) 5x+13` là bội của `2x+1`
`<=> 5x+13 vdots 2x+1`
`<=> 2(5x+13) vdots 2x+1`
`<=> 10x+26 vdots 2x+1`
`<=> 10x+26-5(2x+1) vdots 2x+1`
`<=> 10x+26-10x-5 vdots 2x+1`
`<=> 21 vdots 2x+1`
`<=> 2x+1 in Ư(21)={+-1;+-3;+-7;+-21}`
mà `x>=1`
`-> 2x+1>=3`
`-> 2x+1 in {3;7;21}`
- Ta có bảng sau :
$\begin{array}{|c|c|} \hline 2x+1&3&7&21 \\\hline 2x&2&6&20 \\\hline x&1&3&10 \\\hline \end{array}$
- Vậy `x in {1;3;10}`
