Question

已知x+y=-4,xy=3，求（x+2）/（y+2）+（y+2）/（x+2）的值

Answer 1

(x+y)²=(-4)²
x²+y²+2xy==16
x²+y²=16-2xy=10

原式=[(x+2)²+(y+2)²]/(x+2)(y+2)
=(x²+4x+4+y²+4y+4)/(xy+2x+2y+4)
=[(x²+y²)+4(x+y)+8]/[xy+2(x+y)+4]
=(10-16+8)/(3-8+4)
=-2

Answer 2

(x+2)/(y+2)+(y+2)/(x+2)=[(x+2)^2+(y+2)^2]/[(x+2)(y+2)]
=[(x+2+y+2)^2-2(x+2)(y+2)]/[(x+2)(y+2)],
x+2+y+2=x+y+4=-4+4=0,
(x+2)(y+2)=xy+2(x+y)+4=3+2(-4)+4=-1,
(x+2)/(y+2)+(y+2)/(x+2)=[(x+2+y+2)^2-2(x+2)(y+2)]/[(x+2)(y+2)]
=[0-2(-1)]/(-1)=-2

Answer 3

由已知：
原式=[(x+2)^2+(y+2)^2]/[(x+2)(y+2)]
=[(x+y)^2-2xy+4(x+y)+8]/[xy+2(x+y)+4]
=(16-6-16+8)/(3-8+4)
=-2

Answer 4

-2

Answer 5

（x+2）/（y+2）+（y+2）/（x+2）
=[(x+2)^2+(y+2)^2]/(x+2)(y+2)
(x+2)^2+(y+2)^2
=x^2+4x+4+y^2+4y+4
=x^2+y^2+4(x+y)+8
x+y=-4,xy=3
(x+y)^2=x^2+y^2+2xy=16
x^2+y^2=10
(x+2)^2+(y+2)^2=30
(x+2)(y+2)=xy+2x+2y+4=xy+2(x+y)+4=-1
∴（x+2）/（y+2）+（y+2）/（x+2）
=[(x+2)^2+(y+2)^2]/(x+2)(y+2)
=-30

Answer 6

因(x+2)(y+2)=xy+2(x+y)+4=3-3*4+4=-5
由x+y=-4 得(x+2)+(y+2)=0
平方(x+2)²+(y+2)²+2(x+2)(y+2)=0
所以(x+2)²+(y+2)²=2*5=10
故（x+2）/（y+2）+（y+2）/（x+2）
=[(x+2)²+(y+2)²]/(x+2)(y+2)
=10/(-5)
=-2
希望能帮到你，祝学习进步O(∩_∩)O