Question

求方程组x^2+y^2=1','x*y=2的解的matlab程序（使用solve）

syms x y
[x,y]=solve('x^2+y^2=1','x*y=2')
得不出结果

Answer 1

syms x y
[x,y]=solve('x^2+y^2=1','x*y=2')
运行结果 有四组解（x，y） 可以验证下
x =

((15^(1/2)*i)/2 + 1/2)^(1/2)/2 - ((15^(1/2)*i)/2 + 1/2)^(3/2)/2
((15^(1/2)*i)/2 + 1/2)^(3/2)/2 - ((15^(1/2)*i)/2 + 1/2)^(1/2)/2
(1/2 - (15^(1/2)*i)/2)^(1/2)/2 - (1/2 - (15^(1/2)*i)/2)^(3/2)/2
(1/2 - (15^(1/2)*i)/2)^(3/2)/2 - (1/2 - (15^(1/2)*i)/2)^(1/2)/2

y =

((15^(1/2)*i)/2 + 1/2)^(1/2)
-(1/2*15^(1/2)*i + 1/2)^(1/2)
(1/2 - (15^(1/2)*i)/2)^(1/2)
-(1/2 - 1/2*15^(1/2)*i)^(1/2)

Answer 2

x =

((15^(1/2)*i)/2 + 1/2)^(1/2)/2 - ((15^(1/2)*i)/2 + 1/2)^(3/2)/2
((15^(1/2)*i)/2 + 1/2)^(3/2)/2 - ((15^(1/2)*i)/2 + 1/2)^(1/2)/2
(1/2 - (15^(1/2)*i)/2)^(1/2)/2 - (1/2 - (15^(1/2)*i)/2)^(3/2)/2
(1/2 - (15^(1/2)*i)/2)^(3/2)/2 - (1/2 - (15^(1/2)*i)/2)^(1/2)/2

y =

((15^(1/2)*i)/2 + 1/2)^(1/2)
-(1/2*15^(1/2)*i + 1/2)^(1/2)
(1/2 - (15^(1/2)*i)/2)^(1/2)
-(1/2 - 1/2*15^(1/2)*i)^(1/2)