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 |