设U={0,1,2,3},A={x∈U|x2+mx=0},CUA={1,2},则实数m的值为( ) A.-3 B.-2 C.-1 D.0
问题描述:
设U={0,1,2,3},A={x∈U|x2+mx=0},CUA={1,2},则实数m的值为( )
A. -3
B. -2
C. -1
D. 0
答
∵U={0,1,2,3},A={x∈U|x2+mx=0},CUA={1,2},
∴0、3∈A,
由0+3=-m,可得 m=-3,
故选A.