已知x*2+y*2-8x+10y+41=01、已知x的2次方+y的2次方-8x+10y+41=0,求x/y-y/x的值2、已知x+y=-4,xy=-12,求y+1/x+1 + x+1/y+1的值急

1、
把41拆成16+25
(x²-8x+16)+(y²+10y+25)=0
(x-4)²+(y+5)²=0
平方大于等于0，相加等于0
若有一个大于0，则另一个小于0，不成立。
所以两个都等于0
所以x-4=0,y+5=0
x=4,y=-5
所以x/y-y/x
=4/(-5)-(-5/4)
=9/20
2、
x+y=-4
平方
x²+y²+2xy=16
x²+y²=16-2xy=16+24=40
原式通分=[(y+1)²+(x+1)²]/(x+1)(y+1)
=(x²+y²+2x+2y+2)/(xy+x+y+1)
=[(x²+y²)+2(x+y)+2]/[xy+(x+y)+1]
=(40-8+2)/(-12-4+1)
=-34/15

x^2+y^2-8x+10y+41=0
(x^2-8x+16)+(y^2+10y+25)=0
(x-4)^2+(y+5)^2=0
x-4=0,x=4
y+5=0,y=5
(x/y)-(y/x)
=(x^2-y^2)/(xy)
=(16-25)/20
=-9/20
2.(y+1)/(x+1)+(x+1)/(y+1)
=[(y+1)^2+(x+1)^2]/(x+1)(y+1)
=(y^2+2y+x^2+2x+2)/(xy+x+y+1)
=[(x+y)^2-2xy+2(x+y)+2]/(xy+x+y+1)
=(16+24-8+2)/(-12-4+1)
=-34/15

x*2+y*2-8x+10y+41=0
(x-4)^2+(y+5)^2=0
(x-4)^2=0
x-4=0
x=4
(y+5)^2=0
y+5=0
y=-5
x/y-y/x
=-4/5-(-5/4)
=9/20
x+y=-4,xy=-12
(x+y)^2=x^2+y^2+2xy=16
x^2+y^2=16-2(-12)=40
y+1/x+1 + x+1/y+1
=(y+1)^2+(x+1)^2/(x+1)(y+1)
=[x^2+y^2+2(x+y)+2]/xy+x+y+1
=(40-24+2)/-12-4+1
=18/(-15)
=-1.2

1
x*2+y*2-8x+10y+41=0
(x-4)^2+(y+5)^2=0
所以x=4,y=-5
x/y-y/x=-4/5-(-5/4)=9/20
2
y+1/x+1 + x+1/y+1=(x+y)+(x+y)/xy+2
=-4+(-4)/(-12)+2=-5/3

1、把41拆成16+25(x²-8x+16)+(y²+10y+25)=0(x-4)²+(y+5)²=0平方大于等于0,相加等于0若有一个大于0,则另一个小于0,不成立.所以两个都等于0所以x-4=0,y+5=0x=4,y=-5所以x/y-y/x=4/(-5)-(-5/4)=9/2...

1.x=4,y=5 原式等于-0.45
2.-1-1/6