Solve the inequality ax a 1 x 1 0

ax²－（a＋1）x＋1＜0

ax-1)(x-1)＜0

When a 0, get.

x-1 a) (x-1) is greater than 0

x is greater than 1 or x is less than 1 a

When a is greater than 0, it gets.

x-1/a)(x-1)＜0

When a is greater than 1, 1 a x 1 is obtained

When a is less than 1, 1 x 1 a is obtained

When a=1, getting x is not equal to 1

In summary, when a 0, x is greater than 1 or x is less than 1 a, and when a is greater than 1, 1 a x 1 is obtained

When a is less than 1 and greater than 0, 1 x 1 a is obtained

When a=1, getting x is not equal to 1

When a=0, the solution is x>1;

When a>0, deta=(a-1) 2>=0, when deta=0, the image of the integer quadratic function has no part below the x-axis, so there is no solution less than 0, and when deta>0 solve the equation to obtain x=a+1+-|a-1|, discuss a in different situations to remove the absolute value, and then take the interval between the two (pay attention to the comparison size) is what is sought.

When a<0, similar to a>0 when the discussion, deta=0 when the image is all below (divided by zero), deta>0 to the interval to remove the part of the interval between the two and for the seeking, here is not good to type, can only say the general idea.

Solution: x a-1

When a-1<0, that is, a<1, the section beam has no selling and burning solution.

When a=1, x=0

When a>1, -|a-1|i.e. middle finger 1-a

x - (1+a)x+a 0, which can be reduced to (x-1)(x-a) 0 The two parentheses must be one positive and one negative When a 1, the set of solutions of the above inequalities is When a 1, the set of solutions of the above inequalities is desirable!

a=0 when constant holds.

a≠0, =a -4a<0 There is no intersection with the x-axis, but the opening is upward, and he is always above 0. Therefore.

a (0,4) is constant (this is the wrong place upstairs) when a=4, x≠1 2 (is missing from the upstairs) =a -4a>0.

a∈(-0)∪(4,+∞

a (-0) x (1 2 - a -4a) a, 1 2 + a -4a) a).

a (4,+ x (-1 2 - a -4a) a) (1 2 + a -4a) a,+

a=0 when constant holds.

a≠0, =a -4a<0.

a (0,4) has no real solution.

A -4a>0.

a∈(-0)∪(4,+∞

a (-0) x (1 2 - a -4a) a, 1 2 + a -4a) a).

a (4,+ x (-1 2 - a -4a) a) (1 2 + a -4a) a,+

x²-ax+2＞0

x²-ax＞-2

x²-ax +a²/4＞-2+a²/4

x-a/2)²＞a²-8）/4

When -2 2 a 2 2:

x r when a=-2 2 or a=2 2:

x≠a2 when a-2 2 or a2 2:

x 2, or x 2

x²-(a²+a)x+a³>0

x-a²)(x-a)>0

and a -a = a (a-1).

When a>1, or a<0, and a > a, then the solution set is: x>a, or x0, then the solution set is: x r, and x≠1;

When 0a, or x0, then the solution set is: x r, and x ≠ 0

Factorization i.e. (x-a) (x-a) 0, when a a, i.e., 0 a 1, x a or x a; When a a, x a or x a; When a=a, there is no solution to the inequality. Please adopt, thank you.

ax²-(a+1)x+1>0

When a=0.

It is a 1st order inequality.

x "Base 1 when a≠0.

is a 2nd degree inequality.

ax²-(a+1)x+1>0

ax-1)(x-1)>0

When 1 a> 1, i.e. 01 a}

When 1 a=1, that is, a=1

x-1)²>0

The x≠1 solution set is.

When 1 a<1, i.e., a>1 or a<0

The solution set is. In summary.

a>1 or a<0, the solution set is.

When a=1, the solution set is.

0 nuclear frontal pei 1 a}

When a=0, the solution set is.

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Hello: ax squared (a 1) x 1 0

ax-1）（x-1）<0

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x²－x－a(a＋1)>0

x＋a)(x－a－1)>0

x1=－a，x2=a＋1

1) If x1 = x2, i.e., a= 1 2, then the solution set is:

2) If x1>x2, i.e., a< 1 2, then the solution is: