设a+b=1,a²+b²=2,求a的三次方+b的三次方的值；已知x²-6x+y²+10y+34=0,求x、y的值

Question

设a+b=1,a²+b²=2,求a的三次方+b的三次方的值； 已知x²-6x+y²+10y+34
=0,求x、y的值

Answer 1

a+b=1
(a+b)^2=1
a^2+b^2+2ab=1
2+2ab=1
2ab=-1
ab=-1/2
a^3+b^3
=(a+b)(a^2-ab+b^2)
=1*(2-ab)
=2-ab
=2-(-1/2)
=2+1/2
=5/2
x2+y2-6x+10y+34=0
(x-3)²+(y+5)²=0
x=3,
y=-5