已知（X+2）^2+|y-1|=0,求多项式7X^2y-3+2xy^2-6x^2y-2xy^2+4的值

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已知（X+2）^2+|y-1|=0,求多项式7X^2y-3+2xy^2-6x^2y-2xy^2+4的值
已知（X+2）^2+|y-1|=0,求多项式7X^2y-3+2xy^2-6x^2y-2xy^2+4的值

已知（X+2）^2+|y-1|=0,求多项式7X^2y-3+2xy^2-6x^2y-2xy^2+4的值
平方和绝对值都大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立.
所以两个都等于0
所以 X+2=0,y-1=0
X=-2,y=1
原式=x^2y+1=4+1=5

0<=x+a<=1
0<=x-a<=1
即-a<=x<=-a+1
a<=x<=a+1
若a+1<-a或-a+1a<-1/2,a>1/2
若-a<=a<=-a+1<=a+1
则a>=0
且a<=-a+1,a<=1/2
即0<=a<=1/2,此时是a<=x<=-a+1
若a<=-a<=a+1<=-a+1...

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0<=x+a<=1
0<=x-a<=1
即-a<=x<=-a+1
a<=x<=a+1
若a+1<-a或-a+1a<-1/2,a>1/2
若-a<=a<=-a+1<=a+1
则a>=0
且a<=-a+1,a<=1/2
即0<=a<=1/2,此时是a<=x<=-a+1
若a<=-a<=a+1<=-a+1
则a<=0
且-a<=a+1,a>=-1/2
即-1/2<=a<=0,此时是-a<=x<=a+1
所以
a<=-1/2，空集
-1/2<=a<=0,[-a,a+1]
0<=a<=1/2,[a,-a+1]
a>=1/2，空集

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