已知A={x|x2+x-2=0},B={x|mx+1=0},且A∪B=A,求实数m的取值范围.
2个回答
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A={x|x2+x-2=0}
={x| (x+2)(x-1)=0}
= {-2,1}
A∪B=A
=> B is subset of A
when x =-2
mx +1 =0
-2m +1 =0
m = 1/2
when x =1
mx +1 =0
m +1 =0
m = -1
ie m= -1 or 1/2
={x| (x+2)(x-1)=0}
= {-2,1}
A∪B=A
=> B is subset of A
when x =-2
mx +1 =0
-2m +1 =0
m = 1/2
when x =1
mx +1 =0
m +1 =0
m = -1
ie m= -1 or 1/2
追问
x 不能={-2，1}吗
追答
A= {-2.1}
x 不是一个集合， x是集合里面的元素
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