| x2+cosx+1−sinx |
| x2+cosx+1 |
| sinx |
| x2+ cosx+1 |
令g（x）=
| sinx |
| x2+cosx+1 |
g(−x)＝
| sin(−x) |
| (−x)2+cos(−x)+1 |
∴函数g（x）为奇函数，图象关于原点对称，最大值与最小值也关于原点对称，即函数g（x）的最值的和为0
∵f（x）=1-g（x）
∴M+m=1-g（x）min+1-g（x）max=2
故答案为：2
| x2+cosx−sinx+1 |
| x2+cosx+1 |
| x2+cosx+1−sinx |
| x2+cosx+1 |
| sinx |
| x2+ cosx+1 |
| sinx |
| x2+cosx+1 |
| sin(−x) |
| (−x)2+cos(−x)+1 |