初三二次函数难题(高手进)

1.知x=c为方程ax^2+bx+c=0的一根 可求得ac+b+1=0
所以抛物线y=ax^2+bx+c与x轴至少有一个交点x=c

考虑2种情况
1)抛物线与x轴有两个交点
根据已知a>0抛物线开口向上 又知当0<x<c时,y>0
作抛物线图很容易得到 另一个交点x2大于c
由韦达定理x1x2=c/a 得另一交点 x2=c/a/c=1/a
1/a>c得ac<1
2)抛物线与x轴只有有1个交点x=c
同样满足a>0抛物线开口向上 当0<x<c时,y>0
此时由韦达定理x1x2=c^2=c/a ac=1
所以由1) 2)可得ac<=1

2.
ac<=1 c>1
所以a<1
a/(x+2)+b/(x+1)+c/x
=a/(x+2)-(ac+1)/(x+1)+c/x
=a/(x+2)+c/x-(ac+1)/(x+1)
=[ax+c(x+2)]/[x(x+2)] -(ac+1)/(x+1)
=[(a+c)x+2c]/[x(x+2)]-(ac+1)/(x+1)
>[(a+c)x+2c]/[x(x+2)+1]-(ac+1)/(x+1)
因为[(a+c)x+2c]/[x(x+2)+1]-(ac+1)/(x+1)
=[(a+c)x+2c]/(x+1)^2-(ac+1)(x+1)/(x+1)^2
=[（a+c)x+2c-(ac+1)x-ac-1]/(x+1)^2
=[(a+c-ac-1)x+(2c-ac-1)]/(x+1)^2
=[(1-a)(c-1)x+(c-ac)+(c-1)]/(x+1)^2
a<1 c>1 x>0
所以(1-a)(c-1)x>0 c-ac>0 c-1>0
所以[(a+c)x+2c]/[x(x+2)+1]-(ac+1)/(x+1)>0
所以
a/(x+