A元素是x,即定义域所以-x²+3x+10>=0(x-5)(x+2)<=0-2<=x<=5A并B=AB是A的子集B是m+1<=x<=2m-1若B是空集则m+1>2m-1m<2若B不是空集m>=2此时B的区间在A的区间内所以-2<=m+1<=x<=2m-1<=5-2<=m+1,m>=-32m-1<=5,m<=3所以2<=m<=3综上m≤3