Question

请问一下,如果y=4x*x*x+3x*x+x+2x+1,怎么用MATLAB实现,x等于多少y

谢谢各位了!
我只有这么点分数,拜托大家了

Answer 1

>> x=solve('y=4*x*x*x+3*x*x+x+2*x+1','x')

x =

1/4*(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)-3/4/(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)-1/4
-1/8*(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)+3/8/(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)-1/4+1/2*i*3^(1/2)*(1/4*(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)+3/4/(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3))
-1/8*(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)+3/8/(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)-1/4-1/2*i*3^(1/2)*(1/4*(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)+3/4/(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3))

即
x1=1/4*(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)-3/4/(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)-1/4

x2= -1/8*(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)+3/8/(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)-1/4+1/2*i*3^(1/2)*(1/4*(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)+3/4/(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3))

x3= -1/8*(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)+3/8/(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)-1/4-1/2*i*3^(1/2)*(1/4*(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3)+3/4/(-3+8*y+2*(9-12*y+16*y^2)^(1/2))^(1/3))