a) Ta có: x*y=9
⇔x,y∈Ư(9)
⇔x,y∈{1;-1;3;-3;9;-9}
*Trường hợp 1:
\(\left\{{}\begin{matrix}x=1\\y=9\end{matrix}\right.\)(tm x<y)
*Trường hợp 2:
\(\left\{{}\begin{matrix}x=9\\y=1\end{matrix}\right.\)(ktm vì x>y)
*Trường hợp 3:
\(\left\{{}\begin{matrix}x=-1\\y=-9\end{matrix}\right.\)(ktm vì x<y)
*Trường hợp 4:
\(\left\{{}\begin{matrix}x=-9\\y=-1\end{matrix}\right.\)(tm x<y)
*Trường hợp 5:
\(\left\{{}\begin{matrix}x=3\\y=3\end{matrix}\right.\)(ktm vì x=y)
*Trường hợp 6:
\(\left\{{}\begin{matrix}x=-3\\y=-3\end{matrix}\right.\)(ktm vì x=y)
Vậy: (x,y)={(1;9); (-9;-1)}
b) Ta có: \(\left(x-6\right)\left(y+2\right)=7\)
⇔x-6; y+2∈Ư(7)
⇔x-6; y+2∈{1;-1;7;-7}
*Trường hợp 1:
\(\left\{{}\begin{matrix}x-6=1\\y+2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=-5\end{matrix}\right.\)
*Trường hợp 2:
\(\left\{{}\begin{matrix}x-6=7\\y+2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=13\\y=-1\end{matrix}\right.\)
*Trường hợp 3:
\(\left\{{}\begin{matrix}x-6=-1\\y+2=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=-9\end{matrix}\right.\)
*Trường hợp 4:
\(\left\{{}\begin{matrix}x-6=-7\\y+2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)
Vậy: (x,y)={(7;-5); (13;-1); (5;-9); (-1;-3)}
