设U={0,1,2,3},A={x属于U|x^2+mx=0},若CuA={1,2},则实数m=?
设U={0,1,2,3},A={x属于U|x^2+mx=0},若CuA={1,2},则实数m=?
设U={0,1,2,3}.A={x∈U|x+mx=0},若CuA={1,2}.则实数m为多少?
设U={0,1,2,3},A={x∈U|x2+mx=0},CUA={1,2},则实数m的值为( )
设U={0,1,2,3},A={x属于U|x2+mx=0},若CRB(补集)={1,2},则实数m=
设全集U={1,2,3,4},非空集合A={x丨x-5x+m=0,x属于U},求CuA及实数m的值
已知U={1,2,3,4},A={x|x^2-5x+m=0,x属于U},求CuA、m
设U全集={1,2,3,4},且A={x|x2-7x+m=0,x属于U},若CuA={1,2},求m的值
设全集U={1,2,3,4},且A={x|x2-5x+m=0,x属于U}若CuA={1,4},求m的值.
全集u={1,2,3,4},A={X属于U|X^2-5X+M=0},求CuA和m的值
1.已知全集U={1,2,3,4},A={xlx^2-5x+m,x=0,x属于U},求CuA,m
已知全集U={1,2,3,4}A={X/X平方_5X+m=0 X属于U} 求CuA m
全集U={1,2,3,4,5}A={x|x平方-5qx+4=0,x属于U}若CuA中