Question

x+y+z=1,x2+y2+z2=3则z的范围

Answer 1

x+y+z=1
所以x+y=1-z
x^2+y^2+z^2=3
x^2+y^2=3-z^2
所以
xy=(1/2)[(x+y)^2-(x^2+y^2)]=(1/2)[(1-z)^2-(3-z^2)]
=z^2-z-1
所以x,y是关于t的二次方程t^2-(1-z)t+z^2-z-1=0等两个根
所以Δ=(-(1-z))^2-4(z^2-z-1)>=0
解得-1<=z<=5/3

Answer 2

x+y+z=5
x^2+y^2+z^2=9
可得：xy+xz+yz=8
xy+x(5-x-y)+y(5-x-y)=8
xy+5x-x^2-xy+5y-xy-y^2=8
x^2-5x+xy-5y+y^2+8=0
x^2+(y-5)x+(y^2-5y+8)=0
因为x是实数，则：（y-5)^2-4(y^2-5y+8)>=0
y^2-10y+25-4y^2+20y-32>=0
3y^2-10y+7<=0
y的取值范围为：[1,7/3]
同理x,z的取值范围亦为[1,7/3]