Question

设实数a、b,分别满足19a^2+99b+1=0,b^2+99b+19=0.并且ab≠1,（ab+4a+1)÷b的值

Answer 1

将19+99b+b^2=0,除以b^2,得1+99(1/b)+19(1/b)^2=0.可知,a,1/b是19x^2+99x+1=0的两不等根,(若相等ab=1,矛盾),由韦达定理,a+1/b=-99/19,a/b=1/19.(ab+4a+1)/b=a+1/b+4a/b=-99/19+4*1/19==-5