x^2-mx-2m≤0有实数解,得 x^2-mx-2m=0有实数解
则 判别式=m^2+8m≥0 且 两根和为m,积为-2m
对于任意地实数解x1、x2恒有|x1-x2|≤3,得
0≤(x1-x2)^2=(x1+x2)^2-4x1x2≤m^2+8m≤9
解,得 m∈[-9,-8]∪[0,1]