Question

x+y+z=4 xy+yz+xz=4求x2+y2+z2的解 已知x2+y2+z2-2x+4y-6z+14=0求x+y+z的解求过程，忘记怎么做了

Answer 1

1.
x^2 + y^2 +z^2 = (x + y+ z) ^2 - 2*( xy+yz+xz) = 8

2.
配方
(x-1)^2 + (y+2)^2 + (z-3)^2 =0
x =1
y =-2
z = 3
x+y+z = 2

追问

兄弟我敬佩你，如果你能帮我把作业全做完我给你100财富撒

Answer 2

x+y+z=4 xy+yz+xz=4求x2+y2+z2的解
x2+y2+z2
=(x+y+z)^2-2(xy+yz+xz)
=4^2-2x4
=8
已知x2+y2+z2-2x+4y-6z+14=0求x+y+z的解
x2+y2+z2-2x+4y-6z+14=0
x^2-2x+1+y^2+4y+4+z^2-6z+9=0
(x-1)^2+(y+2)^2+(z-3)^2=0
所以：x=1,y=-2,z=3
x+y+z=1-2+3=2