Question

已知 Ix-y+1I 与x²+8x+16互为相反数，求x²+2xy+y²

Answer 1

Ix-y+1I +x²+8x+16=0
Ix-y+1I +(x+4)^2=0
x=-4
x-y+1=0
y=-3

x²+2xy+y²
=(x+y)^2
=49

Answer 2

(x+4)²=x²+8x+16 则x=0 Ix-y+1I=0 y=1
x²+2xy+y² =1

Answer 3

x2+8x+16>=0所以要使 Ix-y+1I 与x2+8x+16互为相反数，.则x2+8x+16=0，则x=-4,所以Y=-3，所以答案为49。