a. (x-1) (2y+3) = 5
=> 2y+3\(\in U\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta co bang sau:
|
2y+5 |
-1 | 1 | -5 | 5 |
| x-1 | -5 | 5 | -1 | 1 |
| 2y | -6 | -4 | -10 | 0 |
| x | -4 | 6 | 0 | 2 |
| y | -3 | 2 | -5 | 0 |
Vay (x;y)\(\left\{\left(-4;-3\right);\left(6;2\right);\left(0;-5\right);\left(2;0\right)\right\}\)
b.\(|x|=5\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
Vay \(x\in\left\{5;-5\right\}\)
\(\left|y\right|=7\)
\(\Rightarrow\left[{}\begin{matrix}y=7\\y=-7\end{matrix}\right.\)
Vay \(y\in\left\{7;-7\right\}\)
