Question

一道三元二次方程组问题已知x^2+xy+y^2=49y^2+yz+z^2=36z^2+zx+x^2=25求x+y+z.

Answer 1

x^2+xy+y^2=49 (1)

y^2+yz+z^2=36 (2)

z^2+zx+x^2=25 (3)

(1)-(2)得x^2-z^2+y(x-z)=13

(x+y+z)(x-z)=13

设x+y+z=A

则A=13/(x-z)

同理可得A=11/(y-x)

A=-24/(z-y)

因为(x-z)+(y-x)+(z-y)=0

所以13/A+11/A+=-24/A=0

A=7/6

祝你学习天天向上,加油!