设全集S={3,4,3-a^2},A={3,a^2-a+2},CsA={-1},求实数a组成的集合THX SO MUCH
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设全集S={3,4,3-a^2},A={3,a^2-a+2},CsA={-1},求实数a组成的集合THX SO MUCH
设全集S={3,4,3-a^2},A={3,a^2-a+2},CsA={-1},求实数a组成的集合
THX SO MUCH
设全集S={3,4,3-a^2},A={3,a^2-a+2},CsA={-1},求实数a组成的集合THX SO MUCH
∵CsA={-1},∴3-a^2=-1,解得a=±2
⑴当a=2时,a^2-a+2=2^2-2+2=4,符合题意,故可取;
⑵当a=-2时,a^2-a+2=(-2)^2-(-2)+2=8,不属于集合S,故舍去.
∴实数a组成的集合为{2}
A={3,a^2-a+2},CsA={-1},
所以:有:
3-a^2=-1 得:a^2=4 即:a=2,或-2
a^2-a+2=4 得:a=2 或a=-1
综上可得:a=2
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设全集S={3,4,3-a^2},A={3,a^2-a+2},CsA={-1},求实数a组成的集合THX SO MUCH
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