解:令Z=|x+y|+|x-y|,容易知道Z>=0的,所以将等式两边平方,得
Z^2=(|x+y|+|x-y|)^2=(|x+y|)^2+(|x-y|)^2+2|x+y||x-y|
将上式展开得:Z^2=2(x^2+y^2)+2|x^2-y^2|
1.当x^2>y^2,Z^2=2x^2
2.当x^2<y^2,Z^2=2y^2
3.当x^2=y^2,Z^2=2(x^2+y^2)
因为|x|<=1,|y|<=1,所以的0<=x^2<=1,0<=y^2<=1
所以得知当x^2=y^2=1时,Z^2取得最大值为4,即Z的最大值为2
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