If inequality (k ^ 2-1) x ^ 2 - (k-1) X-1

If inequality (k ^ 2-1) x ^ 2 - (k-1) X-1

The answer should be - 3 / 5 < K ≤ 1
Don't forget k = 1
Because the problem is not quadratic inequality, so do this kind of problem
When the quadratic coefficient is 0, it should be considered first,
That is, K ^ 2-1 = 0. The solution is k = ± 1 when k = - 1
Then, the original inequality becomes - 2x - 1 < 0
All real numbers x can't hold, so it doesn't conform to the meaning of the title
When k = 1, the original inequality becomes - 1 < 0
The number x is constant, so it is consistent with the meaning of the title
2582 99054 - vice president level 11 is the same
Finally, take the union of the two cases
That is - 3 / 5 < K ≤ 1

If the inequality ax + B-1, then AB satisfies the following conditions
ax+b

But I don't understand this step
"The solution set is x > - 1
Unequal sign direction change
So A0, that is, a is a positive number, x < - B / A, and the solution set of X is x > - 1
① (2) if a - 1, then - B / a = - 1, that is, B / a = 1, then a = B

Solving the inequality x ^ 2-ax + a ≥ 0 with parameters

(1) X ^ 2-ax + a = 0, discuss △ = a ^ 2 - 4A,
A ∈ (- ∞, 0] ∪ [4, + ∞), x = (a ± √ a ^ 2 - 4A) / 2
△0
X ^ 2-ax + a = (x-a / 2) ^ 2 - A ^ 2 / 4 + a > 0, that is, the lowest point of quadratic equation with one variable is greater than 0,
So a - A ^ 2 / 4 > 0, a ∈ (0,4)

To solve the inequality ax ^ 2 - (a + 1) x + 1 〈 0 with letter parameter a about X
Pay attention to the discussion of letters. I mainly want to see how to classify the discussion

First of all, a > 0, a