实数m,n分别满足方程19m^2+99m+1=0和19+99n+n^2=0,求代数值mn+4m+1/n的值

题目

实数m,n分别满足方程19m^2+99m+1=0和19+99n+n^2=0,求代数值mn+4m+1/n的值
十分钟之内

答案

19+99n+n^2=0两边除以n^2得
19(1/n)^2+99/n+1=0
与19m^2+99m+1=0比较得
m,1/n是方程19x^2+99x+1=0的两根
故
m+1/n=-99/19
m/n=1/19
(mn+4m+1)/n
=m+1/n+4m/n
=-99/19+4/19
=-5