For any real number a, the solution set of inequality ax ^ 2 + AX-1 > 0 is an empty set 1) For any real number a, the solution set of inequality ax ^ 2 + AX-1 > 0 is empty, and the value range of a is obtained. 2) The inequality ax ^ 2 + AX-1 > 0 holds on a ∈ [1,2], and the value range of real number x is obtained. 3) Given that f (x) = 3 ^ (2x) - (K + 1) 3 ^ x + 2, when x ∈ R, f (x) is always positive, find the value range of K. Especially the second question. I completely forgot how to solve orz

For any real number a, the solution set of inequality ax ^ 2 + AX-1 > 0 is an empty set 1) For any real number a, the solution set of inequality ax ^ 2 + AX-1 > 0 is empty, and the value range of a is obtained. 2) The inequality ax ^ 2 + AX-1 > 0 holds on a ∈ [1,2], and the value range of real number x is obtained. 3) Given that f (x) = 3 ^ (2x) - (K + 1) 3 ^ x + 2, when x ∈ R, f (x) is always positive, find the value range of K. Especially the second question. I completely forgot how to solve orz

1）a∈[-4,0]
2) X ＜ (- 1 - √ 5) / 2 or X ＞ (- 1 + √ 5) / 2
3） k
4ab-c^2
It is known that the inequality 2x + 1 / (x-a) ^ 2 is greater than or equal to 7 if x belongs to (a, positive infinity), then the minimum value of real number a
Because x > A, so: 2x + 1 ≥ 7 (x-a) square, after sorting out: 7 * x square - (14a + 2) x + 7 * a square - 1 ≤ 0, to make this inequality meaningful, we must make: ⊿ = [- (14a + 2)] square - 4 * 7 * (7 * a square - 1) ≥ 0, solve this inequality: a ≥ - 4 / 7, so the minimum value of a is - 4 / 7
It is known that the solution set of quadratic inequality ax ^ 2 + BX + 1 > 0 is {x | - 2
To solve this problem, we must know the relationship between the solution set of quadratic inequality of one variable and the root of quadratic equation of one variable, which is the key and difficult point of mathematics in senior one. We must master and apply it skillfully. Generally, when a > 0, combined with the image of quadratic function of one variable f (x) = ax ^ 2 + BX + C, we have the following conclusion: if Δ > 0, quadratic equation of one variable ax ^
-2
If the range of function y = f (x) is [1,3], then the range of function f (x) = 1-2f (x + 3) is ()
A. [-5，-1]B. [-2，0]C. [-6，-2]D. [1，3]
∵ 1 ≤ f (x) ≤ 3, ∵ 1 ≤ f (x + 3) ≤ 3, ∵ - 6 ≤ - 2F (x + 3) ≤ - 2, ∵ - 5 ≤ f (x) ≤ - 1
It is known that the image of quadratic function y = AX2 + BX + C (a is not equal to 0) has a common point with the line y = 25, and the solution set of quadratic inequality AX2 + BX + C > 0 is (- 0.5,1|3). The value range of real numbers a, B, C is obtained
In the dead of night, I'll give you an idea
According to the conditions given by the title, draw the picture first, and you will find that A0, opening downward, according to AX2 + BX + C > 0, the solution set is (- 0.5,1|3), bring y = 0 in, according to the root formula, X1 > - 0.5, X2 = 0, then list the inequality, and get the range
This is a universal algorithm for this kind of problems. There are simple ones, but it has to be based on the image. That was 6 or 7 years ago. Sorry, I forgot
PS: I don't know how senior you are. If you have studied trigonometric function, it will be more convenient to do this kind of problem. There will be several solutions
Analysis: ∵ AX2 + BX + C > 0, the solution set is (- 0.5, 1 | 3)
∴a＜0,-1/2+1/3=-1/6=-b/a,
(-1/2)*(1/3)=-1/6=c/a
That is, a = 6B, a = - 6C, a < 0
∵ y = AX2 + BX + C (a is not equal to 0) has a common point with the line y = 25
The solution of ax ^ 2 + BX + C-25 = 0 is real,
△=b^2-4a(c-25）≥0
On behalf of
Analysis: ∵ AX2 + BX + C > 0, the solution set is (- 0.5, 1 | 3)
∴a＜0,-1/2+1/3=-1/6=-b/a,
(-1/2)*(1/3)=-1/6=c/a
That is, a = 6B, a = - 6C, a < 0
∵ y = AX2 + BX + C (a is not equal to 0) has a common point with the line y = 25
The solution of ax ^ 2 + BX + C-25 = 0 is real,
△=b^2-4a(c-25）≥0
Substituting B = A / 6, C = - A / 6
A (a + 144) ≥ 0
∴a≥0,a≤-144
If a ＜ 0, a ≤ - 144
Ψ B = A / 6 ≤ - 24, C = - A / 6 ≥ 24
Firstly, we take - 0.5 and 1 / 3 into AX2 + BX + C = 0 to get the relationship between a and B. then we get the inequality by taking the midpoint of - 0.5 and 1 / 3 into y = AX2 + B + C
Given the quadratic function f (x) and f (2) = - 3, f (- 2) = - 7, f (0) = - 3, (1) find f (x), (2) find the range of F (x) if x ^ 2 + 2x is less than 0
(1) From F (2) = - 3, f (0) = - 3, the symmetry axis X = 1
Let f (x) = a (x-1) & sup2; + B (a ≠ 0)
Take F (0) = - 3, f (- 2) = - 7 into: a = - 1 / 2
∴f（x）=（-1/2）（x-1）²-5/2
（2）∵x^2+2x≤0∴-2≤x≤0
∵ f (x) on [- 2,0] in the range of [- 7, - 3]
Given the quadratic function y = AX2 + BX + C (a ≠ 0), if 2A + B = 0, and when x = - 1, y = 3, then what is the value of y when x = 3?
Substituting x = - 1, y = 3 into y = AX2 + BX + C (a ≠ 0), A-B + C = 3, ∵ 2A + B = 0, ∵ B = - 2A, C = - 3A + 3, ∵ when x = 3, y = 9A + 3B + C = 9a-6a-3a + 3 = 3
The analytic formula of finding f (2x-1) by F (x) = 3x ^ 2-2
f(x)=3x^2-2
f(2x-1)=3（2x-1)^2-2=3(4x^2-4x+1)-2=12x^2-12x+1
That is, f (2x-1) = 12x ^ 2-12x + 1
Hope to help you;
Given the quadratic function f (x) = x square + BX + C and f (x) + 4 = O, the solution set is {x | x = 1}. If the function has zeros in the interval [a, a + 4], write the range of real number a
f(x)+4=0
That is, the solution set of X & # 178; + BX + C + 4 = 0 is {x| x = 1}
The equation has two equal real roots 1
∴b=-2,c+4=1
∴b=-1,c=-3
∴f(x)=x²-2x-3
Let f (x) = 0, that is, X & # 178; - 2x-3 = 0
The solution is x = - 1 or x = 3
If f (x) has no zero point in the interval [a, a + 4]
A + 3 or a + 4
f(x)+4
=x²+bx+(c+4)=0
f(1)+4=0
△=b²-4(c+4)=0
b=-2
c=-3
f(x)=x²-2x-3
f(x)=0
x1=3,x2=-1
Let the function have zeros in the interval [a, a + 4]
(1)
a≤-1=>a+4
If f (2x-1) = x ^ 2-3x + 1, then f (- 2) =?
2x-1=y,x=(1+y)/2,f(y)=(1+y)(1+y)*1/4-3(1+y)/2+1.f(-2)=1/4+3/2+1=11/4
Let 2x-1 = - 2, the solution is: x = - 0.5, substituting into the following: 11 / 4, or you can use the substitution method to find the analytical formula!