Theo đề bài, ta có: \(x+2xy-y=4\)
\(\Rightarrow x\left(1+2y\right)-y=4\)
\(\Rightarrow2x\left(2y+1\right)-2y=8\)
\(\Rightarrow2x\left(2y+1\right)-\left(2y+1\right)=7\)
\(\Rightarrow\left(2y+1\right)\left(2x-1\right)=7\)
Vì \(x,y\in Z\Rightarrow2x-1;2y+1\inƯ\left(7\right)=\left\{\mp1;\mp7\right\}\)
Ta có bảng sau:
| 2x-1 | 1 | -1 | 7 | -7 |
| 2y+1 | 7 | -7 | 1 | -1 |
| x | 1 | 0 | 4 | -3 |
| y | 3 | -4 | 0 | -1 |
Vậy \(\left(x;y\right)\in\left\{\left(1;3\right),\left(0;-4\right),\left(4;0\right),\left(-3;-1\right)\right\}\)
\(x+2xy-y=4\)
\(\Rightarrow2x+2xy-2y=4\)
\(\Rightarrow2x+2y\left(x-1\right)=4\)
\(\Rightarrow2\left[x+y\left(x-1\right)\right]=4\)
\(\Rightarrow x+y\left(x-1\right)=2\)
\(\Rightarrow\left(x-1\right)+y\left(x-1\right)=1\)
\(\Rightarrow\left(x-1\right).\left(1+y\right)=1\)
\(\text{ Bài tớ nhầm ròi }\)
\(\text{Làm lại nà !!}\)
\(2x+4xy-2y=8\)
\(\Rightarrow2x+2y\left(2x-1\right)=8\)
\(\Rightarrow\left(2x-1\right)+2y\left(2x-1\right)=7\)
\(\Rightarrow\left(2x-1\right)\left(1+2y\right)=7\)
\(\text{#Lập_Bảng}\)
⇔ 2x + 4xy − 2y = 8
⇔ 2x (1 + 2y) − (1 + 2y) = 7
⇔ (2x − 1)(2y + 1) = 7
Bạn tự lập bảng ước số để tìm ra kết quả cuối cùng
