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

- Hướng dẫn giải

$$\begin{gathered} \left(\mathrm{x}-3\right)\left(\mathrm{y}-3\right)=9\iff \left[\begin{array}{l}\left\{\begin{array}{l}\left(\mathrm{x}-3\right)=\pm 3\\ \left(\mathrm{y}-3\right)=\pm 3\end{array}\right.\\ \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=\pm 1\\ \left(\mathrm{y}-3\right)=\pm 9\end{array}\right.\\ \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=\pm 9\\ \left(\mathrm{y}-3\right)=\pm 1\end{array}\right.\end{array}\right. \\ \mathrm{TH}1: \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=3\\ \left(\mathrm{y}-3\right)=3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=3+3\\ \mathrm{y}=3+3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=6\\ \mathrm{y}=6\end{array}\right. \\ \mathrm{TH}2: \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=-3\\ \left(\mathrm{y}-3\right)=-3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=-3+3\\ \mathrm{y}=-3+3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=0\\ \mathrm{y}=0\end{array}\right. \\ \mathrm{TH}3: \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=9\\ \left(\mathrm{y}-3\right)=1\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=9+3\\ \mathrm{y}=1+3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=12\\ \mathrm{y}=4\end{array}\right. \\ \mathrm{TH}4: \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=-9\\ \left(\mathrm{y}-3\right)=-1\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=-9+3\\ \mathrm{y}=-1+3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=-6\\ \mathrm{y}=2\end{array}\right. \\ \mathrm{TH}5: \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=1\\ \left(\mathrm{y}-3\right)=9\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=1+3\\ \mathrm{y}=9+3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=4\\ \mathrm{y}=12\end{array}\right. \\ \mathrm{TH}6: \left\{\begin{array}{l}\left(\mathrm{x}-3\right)=-1\\ \left(\mathrm{y}-3\right)=-9\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=-1+3\\ \mathrm{y}=-9+3\end{array}\right.\iff \left\{\begin{array}{l}\mathrm{x}=2\\ \mathrm{y}=-6\end{array}\right. \\ \left(\mathrm{x};\mathrm{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\} \end{gathered}$$