[MATLAB基础] 求助：解最小二乘法的非线性方程组

[gejugeju]

f(1)=log(x(1))-17059.9*x(1)+(1+(1/2+1/25*x(2)/(x(1)+x(2))+4/5*x(2)^2/(x(1)+x(2))^2-6/5*x(2)^3/(x(1)+x(2))^3+4/5*x(2)^4/(x(1)+x(2))^4)*x(2)+0.43*x(3))*(1-x(1))-8.987*x(1)*x(3)-x(2)*(x(2)/(x(1)+x(2)))*(1-(x(2)/(x(1)+x(2))))*(0.08*(x(2)/(x(1)+x(2)))-3.6*(x(2)/(x(1)+x(2)))^2+3.2*(x(2)/(x(1)+x(2)))^3)-log(x(4))-17059.9*x(4)+(1+(1/2+1/25*x(5)/(x(4)+x(5))+4/5*x(5)^2/(x(4)+x(5))^2-6/5*x(5)^3/(x(4)+x(5))^3+4/5*x(5)^4/(x(4)+x(5))^4)*x(5)+0.43*x(6))*(1-x(5))-8.987*x(5)*x(6)-x(5)*(x(5)/(x(4)+x(5)))*(1-(x(5)/(x(4)+x(5))))*(0.08*(x(5)/(x(4)+x(5)))-3.6*(x(5)/(x(4)+x(5)))^2+3.2*(x(5)/(x(4)+x(5)))^3);
f(2)=log(x(2))*x(1)-3966.4*x(2)+(1+0.233*(x(2)/x(1))*(1/2+1/25*x(2)/(x(1)+x(2))+4/5*x(2)^2/(x(1)+x(2))^2-6/5*x(2)^3/(x(1)+x(2))^3+4/5*x(2)^4/(x(1)+x(2))^4)*x(2)+2.09*x(3))*(1-x(2))-0.1*x(2)*x(3)+0.233*(x(2)/x(1))*x(2)*(x(2)/(x(1)+x(2)))*(1-(x(2)/(x(1)+x(2))))*(0.08*(x(2)/(x(1)+x(2)))-3.6*(x(2)/(x(1)+x(2)))^2+3.2*(x(2)/(x(1)+x(2)))^3)-log(x(5))*x(4)-3966.4*x(5)+(1+0.233*(x(5)/x(4))*(1/2+1/25*x(5)/(x(4)+x(5))+4/5*x(5)^2/(x(4)+x(5))^2-6/5*x(5)^3/(x(4)+x(5))^3+4/5*x(5)^4/(x(4)+x(5))^4)*x(5)+2.09*x(6)*(1-x(5)))-0.1*x(5)*x(6)+0.233*(x(5)/x(4))*x(5)*(x(5)/(x(4)+x(5)))*(1-(x(5)/(x(4)+x(5))))*(0.08*(x(5)/(x(4)+x(5)))-3.6*(x(5)/(x(4)+x(5)))^2+3.2*(x(5)/(x(4)+x(5)))^3);
f(3)=log(x(3))*x(1)-5.86*10^(-5)*x(3)+(1+5.86*10^(-5)*(x(3)/x(1))*0.43*x(1)+5.2668*10^(-4)*x(3))*(1-x(3))-5.86*10^(-5)*x(2)*x(3)*(0.5+0.04*(x(2)/(x(1)+x(2)))^2-1.2*(x(2)/(x(1)+x(2)))^3+0.8*(x(2)/(x(1)+x(2)))^4)-log(x(6))*x(4)-5.86*10^(-5)*x(6)+(1+5.86*10^(-5)*(x(6)/x(4))*0.43*x(4)+5.2668*10^(-4)*x(6))*(1-x(6))-5.86*10^(-5)*x(5)*x(6)*(0.5+0.04*(x(5)/(x(4)+x(5)))^2-1.2*(x(5)/(x(4)+x(5)))^3+0.8*(x(5)/(x(4)+x(5)))^4);
f(4)=x(1)+x(2)+x(3)-1;
f(5)=x(4)+x(5)+x(6)-1;
这是我的五个方程。五个方程都=0。目标函数F=(x1-x4)^2+(x2-x5)^2+(x3-x6)^2;
计算使用最小二乘法，通过计算目标函数最小值来求解。
求高手帮助
重谢！