Ta co $A = 1 + \dfrac{2003}{x}$ De $A \in Z$ thi x la uoc cua 2003. Do 2003 la so nguyen to nen $x = \pm 1, \pm 2003$. Vay $A = 2004, -2002, 2, 0$. Trả lời
`A=(x+2003)/x=x/x+2003/x=1+2003/x` `text(để A nguyên khi )2003/xtext( nguyên)` `=>2003vdotsx=>x in Ư_((2003))={+-1:+-2003}` `text( ta có bảng sau)` $\begin{array}{|c|c|}\hline x&-2003&-1&1&2003\\\hline \dfrac{2003}{x}&-1&-2003&2003&1\\\hline A&-2&-2004&2002&0\\\hline \end{array}$ `text(Vậy A )={-2004;-2;0;2002} text( thì )x={-2003;-1;1;2003}` Trả lời
Ta co $A = 1 + \dfrac{2003}{x}$
De $A \in Z$ thi x la uoc cua 2003.
Do 2003 la so nguyen to nen $x = \pm 1, \pm 2003$.
Vay $A = 2004, -2002, 2, 0$.
`A=(x+2003)/x=x/x+2003/x=1+2003/x`
`text(để A nguyên khi )2003/xtext( nguyên)`
`=>2003vdotsx=>x in Ư_((2003))={+-1:+-2003}`
`text( ta có bảng sau)`
$\begin{array}{|c|c|}\hline x&-2003&-1&1&2003\\\hline \dfrac{2003}{x}&-1&-2003&2003&1\\\hline A&-2&-2004&2002&0\\\hline \end{array}$
`text(Vậy A )={-2004;-2;0;2002} text( thì )x={-2003;-1;1;2003}`