\(a)2xy+4y-x=5\)
\(\Leftrightarrow\left(2xy+4y\right)-x=3+2\)
\(\Leftrightarrow2y\left(x+2\right)-x-2=3\)
\(\Leftrightarrow2y\left(x+2\right)-\left(x+2\right)=3\)
\(\Leftrightarrow\left(x+2\right)\left(2y-1\right)=3\)
\(\Rightarrow\left(x+2\right);\left(2y-1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Xét từng trường hợp :
\(\hept{\begin{cases}x+2=1\\2y-1=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}\)\(\hept{\begin{cases}x+2=3\\2y-1=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}}\)\(\hept{\begin{cases}x+2=-1\\2y-1=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=-1\end{cases}}}\)\(\hept{\begin{cases}x+2=-3\\2y-1=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=0\end{cases}}}\)Vậy
\(2x+y=xy-3\)
\(\Leftrightarrow xy-2x-y=3\)
\(\Leftrightarrow\left(xy-2x\right)-y=-2+5\)
\(\Leftrightarrow x\left(y-2\right)-y+2=5\)
\(\Leftrightarrow x\left(y-2\right)-\left(y-2\right)=5\)
\(\Leftrightarrow\left(y-2\right)\left(x-1\right)=5\)
\(\Rightarrow\left(y-2\right);\left(x-1\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Xét các trường hợp như câu trên và kết luận
\(2xy+4y-x=5\)
\(\Rightarrow2y\left(x+2\right)-\left(x+2\right)=5-2\)
\(\Rightarrow\left(x+2\right)\left(2y-1\right)=3\)
\(\Rightarrow\left(x+2\right);\left(2y-1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Ta xét bảng
| x+2 | 1 | -1 | 3 | -3 |
| 2y-1 | 3 | -3 | 1 | -1 |
| x | -1 | -3 | 1 | -5 |
| y | 2 | -1 | 1 | 0 |
Vậy.......................................................
\(2x+y=xy-3\)
\(\Rightarrow2x+y-xy=3\)
\(2x-2+y\left(x-1\right)=3-2\)
\(\Rightarrow2\left(x-1\right)+y\left(x-1\right)=1\)
\(\Rightarrow\left(2+y\right)\left(x-1\right)=1\)
\(\Rightarrow\left(x-1\right);\left(2+y\right)\inƯ\left(1\right)=\left\{\pm1\right\}\)
Xét bảng
| x-1 | 1 | -1 |
| 2+y | 1 | -1 |
| x | 2 | 0 |
| y | -1 | -3 |
Vậy..........................