If x ＜ 1, the inequality ax-2x + 1 ＞ 0 about X holds, the value range of a is obtained

If x ＜ 1, the inequality ax-2x + 1 ＞ 0 about X holds, the value range of a is obtained

It is known that when ax-2x + 1 > 0 for x2x-1 for x0, there is a > (2x-1) / X for 0

Solving inequality 2x ^ 2 + ax + 1

When solving inequality 2x & # 178; + ax + 10, i.e. A2 √ 2:
2(x²+ax/2)+1=2[(x+a/4)²-a²/16]+1=2(x+a/4)²-a²/8+1=2(x+a/4)²-(a²-8)/8

The inequality AX-2 ≥ 2x ax with respect to X

The original inequality can be reduced to AX ^ 2 + (A-2) X-2 ≥ 0 and (AX-2) (x + 1) ≥ 0
(1) When a = 0, the original inequality is reduced to x + 1 ≤ 0 and X ≤ - 1 is deduced;
(2) When a > 0, the original inequality is changed to (X-2 / a) (x + 1) ≥ 0, and then x ≥ 2 / A or X ≤ - 1 is deduced;
(3) When A-1, i.e. a

1. X2 + 2x-3 ≤ 0.2. X-x2 + 6 < 0.3.4x2 + 4x + 1 ≥ 0.4. X2-6x-9 ≤ 05.4 + x-x2 < 0.6. X2 + 2x + 3 < 0

x^2+2x-3

The solution set of inequality 3 2x-x2 greater than or equal to 0 is?

3 + 2x-x ^ 2 > = 0, i.e. x ^ 2-2x-3

The solution of inequality (x2-2x-35) / (X-2) is greater than or equal to 0,

When x ＞ 2, multiply both sides by X-2
x2-2x-35>=0
(x-7)(x+5)>=0
x> X = 7 or X2, so take x > = 7
When x

What is the solution set of inequality x2-2x-3 greater than 0 and inequality-x25x-6 less than 0?
The second inequality should be - x2 + 5x-6 less than 0

Let the two solution sets be a and B respectively,
From x2-2x-3 < 0
Get - 1

The solution of the inequality ax Λ 2 + (AB + 1) x + b > 0 is 1 < x < 2. Find the value of a and B

Then 1 and 2 are the roots of the equation AX & # 178; + (AB + 1) x + B = 0
1 + 2 = - (AB + 1) / A and 1 × 2 = B / A
3A + AB + 1 = 0 and 2A = b
The results are: a = - 1, B = - 2 or a = - 1 / 2, B = - 1

If the solution set of inequality AX2 + (AB + 1) x + b > 0 is {x | 1 ＜ x ＜ 2}, then a + B=______ ．

The solution set of ∵ AX2 + (AB + 1) x + B ＞ 0 is {x | 1 ＜ x ＜ 2}, ∵ a ＜ 0 − AB + 1A = 3ba = 2. The solution is a = − 12b = − 1 or a = − 1b = − 2. ∵ a + B = - 32 or - 3

It is known that the proposition p: function f (x) = 1 / 3x ^ 3 - x ^ 2 + ax + 1 increases monotonically on R, and the proposition q: inequality x ^ 2 + ax + 1 > 0 holds for X ∈ R
If P ∧ q is a false proposition and P ∨ q is a true proposition, the value range of real number a is obtained

Proposition p: the function f (x) = 1 / 3x ^ 3 - x ^ 2 + ax + 1 increases monotonically on R,
T:f'﹙x﹚=x^2－2x+a==>⊿=4-4aa>=1
Proposition q: the inequality x ^ 2 + ax + 1 > 0 holds for X ∈ R
T:⊿=a^2-4-2