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

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

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

tan²15°是多少

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）