Question

x^2+y^2=20,x^2+xy-2y^2=2(x+2y)求x+y的值

Answer 1

x^2+xy-2y^2=2(x+2y)

x^2+xy-2y^2=(x+2y)(x-y)=2(x+2y)

所以x-y=2,两边平方,x^2-2xy+y^2=4

2xy=(x^2+y^2)-4=20-4=16

x^2+y^2=(x+y)^2-2xy=20

(x+y)^2=20+2xy=20+16=36

x+y=±6

Answer 2

x^2+xy-2y^2=2(x+2y)
-> (x-y)(x+2y)=2(x+2y)
-> (x-y) =2
x^2-2xy+y^2=4
x^2+y^2+2xy
=2(x^2+y^2)-(x^2-2xy+y^2)
=2*20-4=36
x+y= 正负6

Answer 3

x^2+xy-2y^2=2(x+2y)
(x^2-y^2)+(xy-y^2)=2(x+2y)
(x+y)(x-y)+y(x-y)=2(x+2y)
(x-y)(x+2y)=2(x+2y)
x-y=2
(x-y)^2=4
x^2+y^2-2xy=4
因为x^2+y^2=20,
得-2xy=-16
由x^2+y^2=20变换得（x+y)^2-2xy=20
则（x+y)^2=36
x+y=±6

Answer 4

x^2+xy-2y^2=2(x+2y)变形得(x+2y)(x-y)=2(x+2y）
则x-y=2