已知x+xy=14,y²+xy=2,求(1)(x+y)²-x²+y² (2)x+y
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已知x+xy=14,y²+xy=2,求(1)(x+y)²-x²+y² (2)x+y
已知x+xy=14,y²+xy=2,求(1)(x+y)²-x²+y² (2)x+y
已知x+xy=14,y²+xy=2,求(1)(x+y)²-x²+y² (2)x+y
已知x ² +xy=14,y²+xy=2,求(1)(x+y)²-x²+y² (2)x+y
(1)(x+y)²-x²+y²
=x²+2xy+y²-x²+y²
=2xy+2y²
=2(xy+y²)
=4
(2)x+y
(x+y)²=x²+2xy+y²
=x²+xy+xy+y²
=14+2
=16
∴x+y=4或x+y=-4