已知x^2+xy=14,xy+y^2=2,(x＞y＞0)

来源：百度知道 编辑：UC知道 时间：2024/05/18 11:17:40
已知x^2+xy=14,xy+y^2=2,(x＞y＞0)
求1. x+y
2. (x+y)^2-x^2+y^2
3. x-y

x^2+xy=14①xy+y^2=2②
①+②,x^2+2xy+y^2=16
(x+y)^2=16
x+y=4

②-①,y^2-x^2=-12
(x+y)^2-x^2+y^2=16-12=4

①-②,x^2-y^2=12
(x-y)(x+y)=12
因为x+y=4
所以x-y=12/4=3

1. (x+y)^2=(x+y)(x+y)=x(x+y)+y(x+y)
=x^2+xy+xy+y^2=14+2=16=4^2
又x＞y＞0故x+y＞0
故x+y=4
2.(x+y)^2-x^2+y^2
=(x+y)(x+y)-(x-y)(x+y)
=2y(x+y)
=2(xy+y^2)=2*2=4
3.4=(x+y)(x+y)-(x-y)(x+y)
故x-y=(x+y)-4/(x+y)=4-4/4=4-1=3