Tìm các số nguyên x, y biết (x-1)(y+2) = 7

* Hướng dẫn giải

$\left(x-1\right)\left(y+2\right)=7\iff \left[\begin{array}{l}\left\{\begin{array}{l}\left(x-1\right)=\pm 1\\ \left(y+2\right)=\pm 7\end{array}\right.\\ \left\{\begin{array}{l}\left(x-1\right)=\pm 7\\ \left(y+2\right)=\pm 1\end{array}\right.\end{array}\right.$

$TH1: \left\{\begin{array}{l}x-1=1\\ y+2=7\end{array}\right.\iff \left\{\begin{array}{l}x=1+1\\ y=7-2\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=5\end{array}\right.$

$TH2: \left\{\begin{array}{l}x-1=-1\\ y+2=-7\end{array}\right.\iff \left\{\begin{array}{l}x=-1+1\\ y=-7-2\end{array}\right.\iff \left\{\begin{array}{l}x=0\\ y=-9\end{array}\right.$

$TH3: \left\{\begin{array}{l}x-1=7\\ y+2=1\end{array}\right.\iff \left\{\begin{array}{l}x=7+1\\ y=1-2\end{array}\right.\iff \left\{\begin{array}{l}x=8\\ y=-1\end{array}\right.$

$TH4: \left\{\begin{array}{l}x-1=-7\\ y+2=-1\end{array}\right.\iff \left\{\begin{array}{l}x=-7+1\\ y=-1-2\end{array}\right.\iff \left\{\begin{array}{l}x=-6\\ y=-3\end{array}\right.$

Vậy $\left(x;y\right)\in \left\{\left(2;5\right);\left(0;-9\right);\left(8;1\right);\left(-6;-3\right)\right\}$