解关于x的不等式ax^2-(a+1)x+1<0

2024-12-26 02:44:06
推荐回答(5个)
回答1:

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

1)当a<0
(x-1)(x-1/a)a<0
(x-1)(x-1/a)>0
x<1/a或 x>1.

2)当a=0
-(x-1)<0
x>1

3)当01/a>1
(x-1)(x-1/a)<0
x<1或 x>1/a.

4)当a=1
(x-1)^2<0
x∈Φ.

5)当a>1
0<1/a<1
(x-1)(x-1/a)<0
1/a
当0回答:因为要先证明x-1是大于零还是小于零,把第二个括号里面的
ax-1变成a(x-1/a),

回答2:

①当a=0时,-x+1<0,得x>1
②当a>0时,ax^2-(a+1)x+1<0
(ax-1)(x-1)<0
方程(ax-1)(x-1)=0的两个解为x1=1/a,x2=1
1)若a>1,则1/a<1,此时不等式的解为1/a<x<1
2)若0<a<1,则1/a>1,此时不等式的解为1<x<1/a
③当a<0时,1/a<1,此时不等式的解集为x>1或x<1/a

注意分类讨论和变号问题!!!

回答3:

ax^2-(a-1)x+1<0
(ax-1)(x-1)<0
若a<0,则两边除a
(x-1/a)(x-1)>0
a<0,1/a<0<1
所以x<1/a,x>1
若a=0,则-(x-1)<0
x>1
若0(x-1/a)(x-1)<0
01
所以1若a=1
则(x-1)^2<0,不成立
无解
若a>1,则两边除a
(x-1/a)(x-1)<0
a>1,1/a<1
所以1/a综上
a<0,x<1/a,x>1
a=0,x>1
0a=1,无解
a>1,1/a

回答4:

做的很棒
但是括号3中应为

3)当01/a>1
(x-1)(x-1/a)<0
1

回答5:

∵ax^2+(1-2a)x+a-1=0
即(x-1)(ax+1-a)=0的根为x1=1,x2=(a-1)/a=1-1/a
当a﹤0时,1-1/a>1
原不等式解集为x﹥1-1/或x﹤1
当a﹥0时,1﹥1-1/a
原不等式解集为1-1/a﹤x﹤1
当a=0时,即x-1﹤0
原不等式解集为x﹤1

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