\(a)xy+3x-7y=21\)
\(\Leftrightarrow xy+3x-7y-21=0\)
\(\Leftrightarrow\left(xy+3x\right)-\left(7y+21\right)=0\)
\(\Leftrightarrow x\left(y+3\right)-7\left(y+3\right)=0\)
\(\Leftrightarrow\left(y+3\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y+3=0\\x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}y=-3\\x=7\end{cases}}}\)
\(b)xy+3x-2y=11\)
\(\Leftrightarrow\left(xy+3x\right)-2y=6+5\)
\(\Leftrightarrow x\left(y+3\right)-2y-6=5\)
\(\Leftrightarrow x\left(y+3\right)-\left(2y+6\right)=5\)
\(\Leftrightarrow x\left(y+3\right)-2\left(y+3\right)=5\)
\(\Leftrightarrow\left(y+3\right)\left(x-2\right)=5\Rightarrow\left(y+3\right);\left(x-2\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Xét các trường hợp :
\(\hept{\begin{cases}y+3=5\\x-2=1\end{cases}\Leftrightarrow\hept{\begin{cases}y=2\\x=3\end{cases}}}\)\(\hept{\begin{cases}y+3=1\\x-2=5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-2\\x=7\end{cases}}\)\(\hept{\begin{cases}y+3=-5\\x-2=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-8\\x=1\end{cases}}\)\(\hept{\begin{cases}y+3=-1\\x-2=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-4\\x=-3\end{cases}}\)
\(xy+3x-7y=21\)
\(\Rightarrow x\left(y+3\right)-7y=21\)
\(\Rightarrow x\left(y+3\right)-7y-21=0\)
\(\Rightarrow x\left(y+3\right)-7\left(y+3\right)=0\)
\(\Rightarrow\left(x-7\right)\left(y+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\y+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\y=-3\end{cases}}}\)
Vậy \(\left(x;y\right)\in\left\{\left(7;-3\right)\right\}\)
\(xy+3x-2y=11\)
\(\Rightarrow x\left(y+3\right)-2y=11\)
\(\Rightarrow x\left(y+3\right)-2y-6=11-6\)
\(\Rightarrow x\left(y+3\right)-2\left(y+3\right)=5\)
\(\Rightarrow\left(x-2\right)\left(y+3\right)=5\)
Vì x;y thuộc Z \(\Rightarrow\left(x-2\right);\left(y+3\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Xét bảng
| x-2 | 1 | -1 | 5 | -5 |
| y+3 | 5 | -5 | 1 | -1 |
| x | 3 | 1 | 7 | -3 |
| y | 2 | -8 | -2 | -4 |
Vậy \(\left(x;y\right)\in\left\{\left(3;2\right);\left(1;-8\right);\left(7;-2\right);\left(-3;-4\right)\right\}\)
a. \(xy+3x-7y=21\)
\(x\left(y+3\right)-7y=21\)
\(x\left(y+3\right)-7y-21=0\)
\(x\left(y+3\right)-7\left(y+3\right)=0\)
\(\left(x-7\right)\left(y+3\right)=0\)
=> \(x-7=0\) hoặc \(y+3=0\)
\(x=7\) \(y=-3\)
b. \(xy+3x-2y=11\)
\(x\left(y+3\right)-2y=11\)
\(x\left(y+3\right)-2y-6=5\)
\(x\left(y+3\right)-2\left(y+3\right)=5\)
\(\left(x-2\right)\left(y+3\right)=5\)
Ta có: \(5=5\cdot1=1\cdot5=\left(-5\right)\cdot\left(-1\right)=\left(-1\right)\cdot\left(-5\right)\)
Ta có bảng sau:
| x | 5 | 1 | -5 | -1 |
| y | 1 | 5 | -1 | -5 |
Vậy \(\left(x;y\right)\varepsilon\left\{\left(5;1\right),\left(1;5\right),\left(-5;-1\right),\left(-1;-5\right)\right\}\)
