Tìm các số nguyên x, y biết (x - 3) (y - 3) = 9

* Hướng dẫn giải

$\left(x-3\right)\left(y-3\right)=9\iff \left[\begin{array}{l}\left\{\begin{array}{l}\left(x-3\right)=\pm 3\\ \left(y-3\right)=\pm 3\end{array}\right.\\ \left\{\begin{array}{l}\left(x-3\right)=\pm 1\\ \left(y-3\right)=\pm 9\end{array}\right.\\ \left\{\begin{array}{l}\left(x-3\right)=\pm 9\\ \left(y-3\right)=\pm 1\end{array}\right.\end{array}\right.$

$TH1: \left\{\begin{array}{l}\left(x-3\right)=3\\ \left(y-3\right)=3\end{array}\right.\iff \left\{\begin{array}{l}x=3+3\\ y=3+3\end{array}\right.\iff \left\{\begin{array}{l}x=6\\ y=6\end{array}\right.$

$TH2: \left\{\begin{array}{l}\left(x-3\right)=-3\\ \left(y-3\right)=-3\end{array}\right.\iff \left\{\begin{array}{l}x=-3+3\\ y=-3+3\end{array}\right.\iff \left\{\begin{array}{l}x=0\\ y=0\end{array}\right.$

$TH3: \left\{\begin{array}{l}\left(x-3\right)=9\\ \left(y-3\right)=1\end{array}\right.\iff \left\{\begin{array}{l}x=9+3\\ y=1+3\end{array}\right.\iff \left\{\begin{array}{l}x=12\\ y=4\end{array}\right.$

$TH4: \left\{\begin{array}{l}\left(x-3\right)=-9\\ \left(y-3\right)=-1\end{array}\right.\iff \left\{\begin{array}{l}x=-9+3\\ y=-1+3\end{array}\right.\iff \left\{\begin{array}{l}x=-6\\ y=2\end{array}\right.$

$TH5: \left\{\begin{array}{l}\left(x-3\right)=1\\ \left(y-3\right)=9\end{array}\right.\iff \left\{\begin{array}{l}x=1+3\\ y=9+3\end{array}\right.\iff \left\{\begin{array}{l}x=4\\ y=12\end{array}\right.$

$TH6: \left\{\begin{array}{l}\left(x-3\right)=-1\\ \left(y-3\right)=-9\end{array}\right.\iff \left\{\begin{array}{l}x=-1+3\\ y=-9+3\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=-6\end{array}\right.$

Vậy $\left(x;y\right)\in \left\{\left(2;-6\right);\left(6;6\right);\left(0;0\right);\left(12;4\right);\left(-6;2\right);\left(4;12\right)\right\}$