若x^2+y^2+2x-8y+17=0试求（x-3y)^2-(x+3y)^2的值

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/15 00:04:58

x^2+y^2+2x-8y+1+16=0
(x^2+2x+1)+(y^2-8y+16)=0
(x+1)^2+(y-4)^2=0
平方大于等于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
x+1=0
x=-1
y-4=0
y=4
(x-3y)^2-(x+3y)^2
=[x-3y+x+3y](x-3y-x-3y)
=2x×(-6y)
=-12xy=48

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