Question

matlab 解二元二次方程

a^2+(d+1+b)^2=100;b^2+(f+1-a)^2=100;
求a,b， d,f为已知量

Answer 1

[a,b]=solve(a^2+(d+1+b)^2==100,b^2+(f+1-a)^2==100,'a','b')
a =
(2*f - d + (-(d^2 + 2*d + f^2 + 2*f - 398)/(d^2 + 2*d + f^2 + 2*f + 2))^(1/2) + 2*d*(d/2 + (-(d^2 + 2*d + f^2 + 2*f - 398)/(d^2 + 2*d + f^2 + 2*f + 2))^(1/2)/2 + (f*(-(d^2 + 2*d + f^2 + 2*f - 398)/(d^2 + 2*d + f^2 + 2*f + 2))^(1/2))/2 + 1/2) - d^2 + f^2 + f*(-(d^2 + 2*d + f^2 + 2*f - 398)/(d^2 + 2*d + f^2 + 2*f + 2))^(1/2) + 1)/(2*f + 2)
-(d - 2*f + (-(d^2 + 2*d + f^2 + 2*f - 398)/(d^2 + 2*d + f^2 + 2*f + 2))^(1/2) - 2*d*(d/2 - (-(d^2 + 2*d + f^2 + 2*f - 398)/(d^2 + 2*d + f^2 + 2*f + 2))^(1/2)/2 - (f*(-(d^2 + 2*d + f^2 + 2*f - 398)/(d^2 + 2*d + f^2 + 2*f + 2))^(1/2))/2 + 1/2) + d^2 - f^2 + f*(-(d^2 + 2*d + f^2 + 2*f - 398)/(d^2 + 2*d + f^2 + 2*f + 2))^(1/2) - 1)/(2*f + 2)
b =
- d/2 - (-(d^2 + 2*d + f^2 + 2*f - 398)/(d^2 + 2*d + f^2 + 2*f + 2))^(1/2)/2 - (f*(-(d^2 + 2*d + f^2 + 2*f - 398)/(d^2 + 2*d + f^2 + 2*f + 2))^(1/2))/2 - 1/2
(-(d^2 + 2*d + f^2 + 2*f - 398)/(d^2 + 2*d + f^2 + 2*f + 2))^(1/2)/2 - d/2 + (f*(-(d^2 + 2*d + f^2 + 2*f - 398)/(d^2 + 2*d + f^2 + 2*f + 2))^(1/2))/2 - 1/2

追问

[a,b]=solve(a^2+(d+1+b)^2==100,b^2+(f+1-a)^2==100,'a','b')

直接把这一句输入matlab就可以了吗？为什么我出不来和你一样的结果？

追答

可以有变量干扰吧。
输入 clear all
然后syms a b d f
最后再输入这句话。应该就一样了。

追问

Warning: Values other than 0 or 1 converted to logical 1
> In maple at 78
In solve at 75
Warning: Values other than 0 or 1 converted to logical 1
……

出来这个……