解关于x的不等式ax^2-(a+1)x+1>0(a>0)

2024-11-27 02:36:13
推荐回答(2个)
回答1:

ax^2-(a+1)x+1>0
(ax-1)(x-1)>0

ax-1=0,x-1=0
x1=1/a,x2=1

1/a-1=(1-a)/a>0
0
当0x>1/a,或x<1

当a=1时
x≠1

当a>1时
x>1或x<1/a

回答2:

解:不等式ax^2-(a+1)x+1>0(a>0)中,
△=[-(a+1)]^2-4*a*1
=a^2+2a+1-4a
=a^2-2a+1=(a-1)^2>=0
解之,a>=1或a=<1,因为a>0
所以a>=1或0
(1)当a=1时,不等式为x^2-2x+1>0,
即(x-1)^2>0
解得,x≠1
(2)当a>1时,1/a<1
ax^2-(a+1)x+1>0
(ax-1)(x-1)>0
解得,x>1或x<1/a
(3)当01
ax^2-(a+1)x+1>0
(ax-1)(x-1)>0
解得,x>1/a,或x<1