19m2+99m+1=0且19+99n+n2=0，則（mn+4m+1）/n=?

解：

19+99n+n2=0兩邊除以n2得

19/n2+99/n+1=0

與19m^2+99m+1=0比較得

m,1/n是方程19x2+99x+1=0的兩個根

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

tan215°是多少

sin215° = (1-cos30°)/2 =(2-根號3）/4

sin215°=(1+cos30°)/2 =(2+根號3）/4

tan215° = sin215°/sin215° = (2-根號3）/(2+根號3）