高一数学...
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A=B
∴a²=1,ab=b或a²=b,ab=1
当a²=1,ab=b时
∴a=1,b为任意实数,或a=-1,b=0
a=1时不满足集合元素的互异性,∴舍去
当a=-1,b=0时
A={-1,1,0}
B={1,-1,0}
A=B成立
当a²=b,ab=1时
a=1,b=1
不满足集合元素的互异性,∴舍去
综上可知,a=-1,b=0
∴a²=1,ab=b或a²=b,ab=1
当a²=1,ab=b时
∴a=1,b为任意实数,或a=-1,b=0
a=1时不满足集合元素的互异性,∴舍去
当a=-1,b=0时
A={-1,1,0}
B={1,-1,0}
A=B成立
当a²=b,ab=1时
a=1,b=1
不满足集合元素的互异性,∴舍去
综上可知,a=-1,b=0
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A={a,a^2,ab} B={1,a,b,},A=B
则 a^2=1 或者 ab=1
若a^2=1 ,则a=-1(由集合的互异性,a不能等于1),b=ab=-b,b=0
若ab=1,则a^2=b,ab=a*a^2=a^3=1,a=1,与集合互异性矛盾
故a=-1,b=0
则 a^2=1 或者 ab=1
若a^2=1 ,则a=-1(由集合的互异性,a不能等于1),b=ab=-b,b=0
若ab=1,则a^2=b,ab=a*a^2=a^3=1,a=1,与集合互异性矛盾
故a=-1,b=0
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Solution:
∵A = B,
∴ All elements of Set A are same as all elements of Set B.
Case 1:
a² = 1, a = ±1
ab = b, a = +1, no general way to determine the value of b.
ab = b, a = -1, also no general way to determine the value of b.
So, possible sulotion is a = 1 and b = 0, or a = -1 and b = 0
Case 2:
a² = ab, a = b
ab = 1, a²= 1
a = ±1, b = ±1
So, possble solution one is a = b = 1; solution two is a = b = -1.
Because each element of a set is diffenernt from others,
the final solution is a = -1, b = 0.
∵A = B,
∴ All elements of Set A are same as all elements of Set B.
Case 1:
a² = 1, a = ±1
ab = b, a = +1, no general way to determine the value of b.
ab = b, a = -1, also no general way to determine the value of b.
So, possible sulotion is a = 1 and b = 0, or a = -1 and b = 0
Case 2:
a² = ab, a = b
ab = 1, a²= 1
a = ±1, b = ±1
So, possble solution one is a = b = 1; solution two is a = b = -1.
Because each element of a set is diffenernt from others,
the final solution is a = -1, b = 0.
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