If the solution set of the inequality ax ^ 2 - | x + 1 | + 2A ＜ 0 about X is an empty set, then the value range of a How does parabola y = ax & sup2; - ax + 1 come from?

If the solution set of the inequality ax ^ 2 - | x + 1 | + 2A ＜ 0 about X is an empty set, then the value range of a How does parabola y = ax & sup2; - ax + 1 come from?

The solution set of the inequality ax ^ 2 - | x + 1 | + 2A ＜ 0 is an empty set, then the solution set of the inequality ax ^ 2 - | x + 1 | + 2A ≥ 0 is not an empty set and has solutions. If x ≥ - 1, | x + 1 | = x + 1, ax ^ 2-x-1 + 2A ≥ 0, the two {1 ± √ [1-4a (2a-1)]} / (2a) of the quadratic trinomial ax ^ 2-x-1 + 2a, the premise of the inequality having solutions is: 1-4a (...)
(1) If a is an empty set, then a = 0, - x + 1 | = 0
There are: a > 0, and - 1 + 2A > = 0;
The solution is a > 1 / 2
In conclusion, a > 1 / 2
The solution set of inequality ax ^ 2 - | x + 1 | + 2A < 0 of X is empty,
Then, the solution set of inequality ax ^ 2 - | x + 1 | + 2A ≥ 0 is not empty and has solution,
If x ≥ - 1, | x + 1 | = x + 1,
ax^2-x-1+2a≥0,
Two properties of quadratic trinomial ax ^ 2-x-1 + 2A
{1±√[1-4a(2a-1)]}/(2a),
The premise for the inequality to have a solution is: 1-4a (2a-1) ≥ 0,
8A ^ 2-4a-1... Expand
The solution set of inequality ax ^ 2 - | x + 1 | + 2A < 0 of X is empty,
Then, the solution set of inequality ax ^ 2 - | x + 1 | + 2A ≥ 0 is not empty and has solution,
If x ≥ - 1, | x + 1 | = x + 1,
ax^2-x-1+2a≥0,
Two properties of quadratic trinomial ax ^ 2-x-1 + 2A
{1±√[1-4a(2a-1)]}/(2a),
The premise for the inequality to have a solution is: 1-4a (2a-1) ≥ 0,
8a^2-4a-1≤0,
The two solutions of quadratic trinomial 8A ^ 2-4a-1 are as follows:
[2±√(4+8)]/8=[1±√3]/4,
The solution of the inequality is: (1 - √ 3) / 4 ≤ a ≤ (1 + √ 3) / 4;
Substitute, and then verify whether it can meet the pre setting of X ≥ - 1
The original inequality is changed into:
a{x-{1-√[1-4a(2a-1)]}/(2a)}{x-{1+√[1-4a(2a-1)]}/(2a)}≥0,
If (1 - √ 3) / 4 ≤ a
If a > 0, then the solution set of inequality ax < A is (); if x > 1, then the value range of a is ()
x＜0 a＜0
X
The solution set of the quadratic inequality AX2 + ax + A-1 ＜ 0 with one variable of X is r, and the value range of a is obtained
When a = 0, the inequality becomes - 1 ＜ 0. When a ≠ 0, according to the meaning of the question: a ＜ 0 △ = A2 − 4a (a − 1) ＜ 0, the solution is a ＜ 0
Let the image of quadratic function y = x2-3x-4x intersect with X axis at points a and B, and intersect with y axis at point C, then the area of triangle ABC is
y=x²-3x-4=(x+1)(x-4)
Let y = 0 give x = - 1 or x = 4
Let x = 0 give y = - 4
So a (- 1,0), B (4,0), C (0, - 4) so the area of triangle ABC is s = (1 / 2) * 5 * 4 = 10
y=x2-3x-4x？？？？
Is that the original question?
From y = x ^ 2-3x-4 = (x + 1) (x-4) = 0: a (- 1,0), B (4,0)
If it intersects with y axis at point C, then y = - 4, so C (0, - 4)
So the area of triangle ABC = labllocl / 2 = 5 * 4 / 2 = 10
1. We know that the equation LG ^ 2 (x) + (2a + 4) lgx + 4 = 0 about X has a real number solution, and find the value range of real number a
2. Given that the equation 4 ^ 2 + (a + 2) 2 ^ (x + 1) + 4 = 0 about X has real number solution, find the value range of real number a
Process. Two questions. Thank you!
One
The value of LG (x) is any real number. If lgx is regarded as a whole, the equation has a real number solution and the discriminant is ≥ 0
(2a+4)²-16≥0
a(a+4)≥0
A ≥ 0 or a ≤ - 4
Two
4 ^ (2x) + (a + 2) 2 ^ (x + 1) + 4 = 0. Correct it, your title is wrong
(2^x)²+(2a+4)(2^x)+4=0
2> If 0,2 ^ x is constant > 0, 2 ^ x is regarded as an unknown number, the equation has real roots, the discriminant is ≥ 0, and the solution is a ≥ 0 or a ≤ - 4
The equation has positive real roots. Let two of them be M1 and M2, respectively
m1+m2=-(2a+4)
M1m2 = 4 > 0, the two roots of the equation are positive real roots, and the sum of the two roots is > 0
-(2a+4)>0
2a+4
(1) Let t = lgx,
t²+（2a+4）t+4=0
Δ=（2a+4）²-16≥0
4a²+16a+16-16≥0
a²+4a≥0
a（a+4)≥0
A ≤ - 4 or a ≥ 0
When a = - 4, t = 2, lgx = 2, x = 100,
When a = 0, t = - 2, lgx = - 2, x = 1 / 100
（2）16+（a+2)×2^x×2+4=0
2^x=-6/(a+2)＞0,
a+2＞6,∴a＞4.
If X-Y = 2, xy = 3, then the value of (x-1) (y + 1) is
(x-1)(y+1)
=xy-y+x-1
=xy+(x-y)-1
=3+2-1
=4
(x-1)(y+1)
=xy-（x-y）+1
=3-2+1
=2
How to solve the problem of probability
The definition of probability is a measure of the probability of the occurrence of random events. It is one of the most basic concepts in probability theory. People often say that the probability of someone passing the exam and the probability of something happening are examples of probability
[3x-1] / 2 is equal to [4x + 2] / 5-1?
Multiply both sides by 10
5(3x-1)=2(4x+2)-10
15x-5=8x+4-10
15x-8x=4-10+5
7x=-1
x=-1/7
There is such a problem: "given that the image of quadratic function y = AX2 + BX + C passes through two points a (0, a), B (1, 2)"
And + +, then the symmetry axis of the quadratic function image is a straight line x = 2 "in the question, the + + + + part is the illegible text covered by ink. According to the existing information, add an appropriate condition in the shadow part of the original question, that is, the original proposition is tenable
First, AB two points into the equation a = C 2 = a + B + C, that is 2A + B = 2
The equation of the axis of symmetry is x = - B / 2A. If x = 2, then 4A = - B, so the condition is a = - 1 or B = 4
Given the function f (x) = {lgx, (x ≥ 3 / 2) LG (3-x), (x, 3 / 2), if the equation f (x) = k has no real solution, find the value range of K
lg(3-x),(x
Obviously, f (x) decreases first and then increases, so f (x) > = f (1.5) = LG 1.5. It is easy to know that f (x) has no maximum
F (x) is a function represented by a set???
Is f (x) = {} the largest? Small? Or something else???
I can't understand the title
f(x)>lg(3/2)
K