第1个回答 2013-07-19
(1)
x³(x²-4)>(x²-4)
(x²-4)(x³-1)>0
(x+2)(x-2)(x-1)(x²+x+1)>0
,x²+x+1=(x+1/2)²+3/4>0,所以,
(x+2)(x-2)(x-1)>0
用穿根法:
在数轴上点出三个根,-2,1,2,
图象是波浪线,开始从右上方往左下方开笔,">0" 就是上方的答案,
(-2,1)∪(2,+∞)
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(2)
0∈B=A
0≠x,反之,A={0,y,y}
0≠y,反之,A={0,x,x}
所以0=x+y
y= - x
A={X,-X,0)
B={0,X²,-X²}
{X1=1
{Y1=-1
{X2=-1
{Y2=1
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(3)
(6-x)=-12,-6.-4.-3,-2,2,3,4,6,12
(x-6)=12,6.4.3,2,-2,-3,-4,-6,-12
x=18,12,10.9.8,4,3,2,0,-6
{x|18,12,10.9.8,4,3,2,0,-6}
第2个回答 2013-07-19
x³(x²-4)+4>x²
x³(x²-4)-(x²-4)>0
(x³-1)(x²-4)>0
(x-1)(x²+x+1)(x+2)(x-2)>0
x²+x+1=(x+1/2)²+3/4恒>0
(x-1)(x+2)(x-2)>0
x>2或-2<x<1
由集合元素互异性得x,y≠0
因此只有x+y=0 y=-x
集合变为:A={x,-x,0},B={x²,-x²,0}
x²=x时,-x²=-x
x(x-1)=0 x=0(舍去)或x=1,此时y=-x=-1
x²=-x时,-x²=x
x(x+1)=0 x=0(舍去)或x=-1,此时y=-x=1
综上,得x=1 y=-1或x=-1 y=1
12/(6-x)∈Z,12能被6-x整除
x是自然数,x只能为,0,2,3,4,5,7,8,9,10,12,18
用列举法表示:{0,2,3,4,5,7,8,9,10,12,18}本回答被提问者采纳
第3个回答 2013-07-19
1)x³﹙x²-4﹚+4>x²
x^5-4x³-x²+4>0
x²﹙x³-1﹚-4﹙x³-1﹚>0
﹙x²-4﹚﹙x³-1﹚>0
x²-4>0且x³-1>0或x²-4<0且x³-1<0
解集为﹛x I﹣2<x<1或x>2﹜
2﹚A=B,互异性的x,y都不为0
所以x+y=0,y=-x
集合变为:A={x,-x,0},B={x²,-x²,0}
x²=x时,-x²=-x
x(x-1)=0 x=0(舍去)或x=1,此时y=-x=-1
x²=-x时,-x²=x
x(x+1)=0 x=0(舍去)或x=-1,此时y=-x=1
综上,得x=1 y=-1或x=-1 y=1
3)12/6-x∈Z,所以6-x可取1,2,3,4,6,12
所以x依次为5,4,3,2,0,﹣6
又因为x∈N
所以﹛0,2,3,4,5﹜