解：
19+99n+n2=0两边除以n2得
19/n2+99/n+1=0
与19m^2+99m+1=0比较得
m,1/n是方程19x2+99x+1=0的两个根
于是
m+1/n=-99/19
m/n=1/19
(mn+4m+1)/n
=m+1/n+4m/n
=-99/19+4/19
=-5
sin215° = (1-cos30°)/2 =(2-根号3）/4
sin215°=(1+cos30°)/2 =(2+根号3）/4
tan215° = sin215°/sin215° = (2-根号3）/(2+根号3）
……