| x2+y2+2(x+1)(1−y) |
| x−y+1 |
∴1+cos2(2x+3y−1)＝
| x2+y2+2x+2−2xy−2y |
| x−y+1 |
∴1+cos2(2x+3y−1)＝
| (x−y)2+2(x−y)+2 |
| x−y+1 |
∴1+cos2(2x+3y−1)＝
| (x−y+1)2+1 |
| x−y+1 |
∴1+cos2(2x+3y−1)＝(x−y+1)+
| 1 |
| x−y+1 |
∵(x−y+1)+
| 1 |
| x−y+1 |
| 1 |
| x−y+1 |
1≤1+cos2（2x+3y-1）≤2
故1+cos2(2x+3y−1)＝(x−y+1)+
| 1 |
| x−y+1 |
此时x-y+1=1，即x=y
2x+3y-1=kπ，即5x-1=kπ，x=
| kπ+1 |
| 5 |
xy=x2=
| (kπ+1)2 |
| 25 |
当k=0时，xy取得最小值
| 1 |
| 25 |
故答案为：
| 1 |
| 25 |