a,b是方程x^2-3x-1=0的根,求ab^2+3aa+b=3ab=-1b^2=3b+1ab^2+3a=a(3b+1)+3a=3ab+4a=4a-3a=[3+/-(13)^(1/2)]/2ab^2+3a=6+/-2*(13)^(1/2)-3=3+/-2*(13)^(1/2)