两道数学题---过程!!! 请帮帮我

x^2+4x+1=0,

so x^2+1=-4x

so x+1/x=-4

so (x+1/x)^2=16

so x^2+1/x^2=16-2=14,

若x的2次方+y的2次方+2x-6y+10=0,则xy=?
x^2+y^2+2x-6y+10=(x+1)^2+(y-3)^2=0

so x+1=0, y-3=0
so x=-1,y=3
so xy=-3

1.已知:x的2次方+4x+1=0,则x的2次方+x的-2次方=
x^2+4x+1=0
x+4+1/x=0
x+1/x=-4
x^2+1/x^2=(x+1/x)^2-2=14
2.若x的2次方+y的2次方+2x-6y+10=0,则xy=?
x^2+2x+1+y^2-6y+9=0
(x+1)^2+(y-3)^2=0
x=-1,y=3
xy=-3

很简单，我11都会，你sb把

1.x^2+4x+1=0
x+4+1/x=0
x+1/x=-4
x^2+1/x^2=(x+1/x)^2-2=14
2.x^2+2x+1+y^2-6y+9=0
(x+1)^2+(y-3)^2=0
x=-1,y=3
xy=-3