Vì \(\left(2x+1\right)\left(y-3\right)=12\)
\(\Rightarrow2x+1;y-3\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
Vì \(2x+1\) là số lẻ nên \(2x+1\in\left\{-3;-1;1;3\right\}\)
Ta có bảng sau:
| 2x+1 | -3 | -1 | 1 | 3 |
| 2x | -4 | -2 | 0 | 2 |
| x | -2 | -1 | 0 | 1 |
| y-3 | -4 | -12 | 12 | 4 |
| y | -1 | -9 | 15 | 7 |
Vậy \(\left(x;y\right)\in\left\{\left(-2;-1\right);\left(-1;-9\right);\left(0;15\right);\left(1;7\right)\right\}\)
Ta có:
\(xy+3x-7y=21\)
\(\Rightarrow x.\left(y+3\right)-7y-21=21-21=0\)
\(x\left(y+3\right)-\left(21+7y\right)=0\)
\(x.\left(y+3\right)-7.\left(y+3\right)=0\)
\(\left(x-7\right)\left(y+3\right)=0\)
\(\Rightarrow x-7=0\) hoặc \(y+3=0\)
TH1: x-7=0
x=0+7=7
TH2:y+3=0
y=0-3=-3
Vậy x=7; y=-3
\(xy-3x-2y=11\)
\(\Rightarrow x\left(y-3\right)-2y+6=11+6=17\)
\(\Rightarrow x\left(y-3\right)-\left(2y-6\right)=17\)
\(\Rightarrow x\left(y-3\right)-2.\left(y-3\right)=17\)
\(\Rightarrow\left(x-2\right)\left(y-3\right)=17\)
\(\Rightarrow x-2;y-3\inƯ\left(17\right)=\left\{-17;-1;1;17\right\}\)
Ta có bảng:
| x-2 | -17 | -1 | 1 | 17 |
| x | -15 | 1 | 3 | 19 |
| y-3 | -1 | -17 | 17 | 1 |
| y | 2 | -14 | 20 | 4 |
Vậy \(\left(x;y\right)\in\left\{\left(-15;2\right);\left(1;-14\right);\left(3;20\right);\left(19;14\right)\right\}\)
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