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

2025-01-24 17:56:01

推荐回答（1个）

回答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