若方程m^2x^2-(2m-3)x+1=0 的两个实数根的倒数之和为S，求S的取值范围

有两个根是二次方程m不等于0
判别式大于等于0
所以(2m-3)^2-4m^2>=0
-12m+9>=0
m<=3/4,且m不等于0
x1+x2=(2m-3)/m^2,x1x2=1/m^2
1/x1+1/x2=(x1+x2)/x1x2=2m-3
m<=3/4,2x<=3/2,2m-3<=-3/2
m=0，2m-3=-3

所以 S<=-3/2且不等于-3

依据题意，
判别式=(2m-3)^2-4m^2
=9-12m>=0
m<=3/4，m不为零
S=(x1+x2)/x1x2
=2m-3
S<=-1.5，且S不等于-3