Question

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

Answer 1

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

Answer 2

0<=x+a<=1
0<=x-a<=1
即-a<=x<=-a+1
a<=x<=a+1

若a+1<-a或-a+1<a,则交集是空集
a<-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