\(\dfrac{1}{x}+\dfrac{2}{y}=1\\ \Rightarrow\dfrac{y}{xy}+\dfrac{2x}{xy}=1\\ \Rightarrow\dfrac{y+2x}{xy}=1\\ \Rightarrow y+2x=xy\)(*)
Giả sử \(x>y\)
\(\Rightarrow y+2x=xy< x+2x=3x\Rightarrow y< 3\)
Mà \(y\in N\)*⇒\(y\in\left\{1;2\right\}\)
với y=1 thay vào (*) ta có:
\(y+2x=xy\\ \Rightarrow1+2x=x\\ \Rightarrow x=-1\)
với y=2 thay vào (*) ta có:
\(y+2x=xy\\ \Rightarrow2+2x=2x\\ \Rightarrow0=2\left(vô.lí\right)\)
Vậy \(\left(x,y\right)\in\left\{\left(-1;1\right)\right\}\)
\(y+2x=xy\\ \Rightarrow y-xy+2x=0\\ \Rightarrow y-x\left(y-2\right)=0\\ \Rightarrow\left(y-2\right)-x\left(y-2\right)+2=0\\ \Rightarrow\left(1-x\right)\left(y-2\right)=-2\)
Ta có bảng:
| 1-x | 1 | 2 | -1 | -2 |
| y-2 | -2 | -1 | 2 | 1 |
| x | 0 | -1 | 2 | 3 |
| y | 0(ktm) | 1(ktm) | 4(tm) | 3(tm) |
Vậy \(\left(x,y\right)\in\left\{\left(2;4\right);\left(3;3\right)\right\}\)
