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.
2019-05-28
一道三元二次方程组问题
已知
x^2+xy+y^2=49
y^2+yz+z^2=36
z^2+zx+x^2=25
求x+y+z.
优质解答
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!
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!