设集合A={x|x2-3x+2=0},B={x|x2+2(a+1)x+(a2-5)=0},若U=R,A⊆CuB,求实数a的取值范围.

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设集合A={x|x2-3x+2=0},B={x|x2+2(a+1)x+(a2-5)=0},若U=R,A⊆CuB,求实数a的取值范围.

设集合A={x|x2-3x+2=0},B={x|x2+2(a+1)x+(a2-5)=0},若U=R,A⊆CuB,求实数a的取值范围.
设集合A={x|x2-3x+2=0},B={x|x2+2(a+1)x+(a2-5)=0},若U=R,A⊆CuB,求实数a的取值范围.

设集合A={x|x2-3x+2=0},B={x|x2+2(a+1)x+(a2-5)=0},若U=R,A⊆CuB,求实数a的取值范围.
由x2-3x+2=0得x=1或x=2
当x=1时,x2+2(a+1)x+(a2-5)=1+2(a+1)+(a2-5)=0解方程得a=-1±√3
当x=2时,x2+2(a+1)x+(a2-5)=4+4(a+1)+(a2-5)=0解方程得a=-1或a=-3
a的取值-1+√3或=-1-√3或-1或-3

X2????????