f(1/x)=(b/x+1)/(2/x+a)=(b+x)/(2+ax)f(x)f(1/x)=(bx+1)(x+b)/[(2x+a)(ax+2)]=k(bx+1)(x+b)=k(2x+a)(ax+2)bx^2+(b^2+1)x+b=2akx^2+(a^2+4)kx+2ak这是恒等式则对应的系数相等b=2akb^2+1=(a^2+4)k所以4a^2k^2+1=(a^2+4...