关于x的方程x2+4

2

x+|m-8|+|m|=0有实根∴△=32-4（|m-8|+|m|）≥0∴|m-8|+|m|≤8

又∵|m-8|+|m|≥|m-8-m|=8

∴|m-8|+|m|=8 ∴0≤m≤8

故p为真时：0≤m≤8

当命题q为真时：f′(x)=3x2+2mx+(m+

10

3

)△=4m2-4×3(m+

10

3

)＞0（8分）∴m²-3m-10＞0

故m＞5或m＜-2                                                        

当命题p且q为真：m∈（5，8]