On the inequality ax + 2 of 2 of X

On the inequality ax + 2 of 2 of X

If you're right, a = 5
(ax+2)/2

The inequality | x + 1 | > A / (x-1) (where a

Unknown positive and negative, classified discussion
If x > 0
x+1>ax-a
x-ax>-a-1
(1-a)x>-a-1
x>1-a/-a-1
If xax-a
-x-ax>-a+1
(-1-a)x>-a+1
x>-1-a/-a+1
The last step of the two questions can be simplified
Come on!
Happy New Year!

How to do ah, high school mathematics solution contains absolute value inequality is always wrong
I know how to change it, but I always miscalculate the numbers. Yesterday's homework was all miscalculated. Now, X-2 > - 1 is going to be > - 3,
What can I do? I can't turn my head

You can use my method! Do inequality, always remember!
1. Items directly related to x, such as - x, 2x, 5 / x, etc., are placed on the left side of the equation! All items related to numbers are placed on the right side!
Example: - 9x + 6

An inequality for the absolute value of X
（1）|x-1|>3x (2) |2x-1|-|3x+1|

(1) When x > = 1, X-1 > 3x, X

Given that the solution set of inequality a ^ 2 + BX + C greater than 0 is x / x = 1 or x = 3, then the solution set of inequality ax ^ 2-bx + C less than 0 is
Given that the solution set of inequality a ^ 2 + BX + C > 0 is {X / X3}, then inequality ax ^ 2-bx + C

The solution set of inequality ax ^ 2 + BX + C > 0 is {X / X3}
We can get that x = 1 and x = 3 are two parts of the equation AX ^ 2 + BX + C = 0
a>0
Then ax ^ 2-bx + C = 0 are x = - 1 and x = - 3
And because a > 0
So ax ^ 2-bx + C

Given that the solution set of inequality AX2 + BX + C > 0 is {x | α < x < β} (β > α > 0 & nbsp;), the solution set of inequality CX2 + BX + a < 0 is obtained

From the known inequality, we can get a < 0, because α and β are two of the equations AX2 + BX + C = 0, so α + β = − Ba & nbsp; & nbsp; & nbsp; & nbsp; ① α· β = Ca & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ② Because a ＜ 0, cax2 + Bax + 1 ＞ 0 is obtained from CX2 + BX + a ＜ 0. Substituting ① and ②, α β X2 - (α + β) x + 1 ＞ 0 is obtained, that is, (α x-1) (β x-1) ＞ 0. Because 0 ＜ α＜ β, 0 ＜ 1 β＜ 1 α. Therefore, the solution set of the inequality is {x | x ＜ 1 β or X ＞ 1 α}

Given the solution set of the inequality ax + BX + C less than 0 {x | x less than - 2 or x greater than - 1 / 2} for X, find the solution set of the inequality ax + BX + C greater than 0

Know ax ^ 2 + BX + C 0 from the title
Also a

Given the set a = {a equation x2-ax + 1 = 0 with respect to x, has real roots}, B = {a inequality ax2-x + 1 > 0 holds for all x r}, find ab
It's urgent,

The set a = {A's equation x2 ax + 1 = 0 with real roots},
△=a^2-4≥0
A ≥ 2 or a ≤ - 2
B = {a inequality ax2-x + 1 > 0 holds for all x r},
a>0,
△=1-4a1/4
A intersection B = {a | a ≥ 2}

The following two propositions are given: proposition p has two unequal real roots with respect to X: x2 MX + 1 = 0, proposition q: inequality x2 MX + 9 > 0 holds when x > 1. If proposition "p if Q" is true and proposition "P and Q" is false, the value range of real number m is obtained

The proposition "p if Q" is true, the proposition "P and Q" is false, and P, q is true and false
If true P: (- M) ^ 2-4 > 0, the solution is m > 2, or if M1, f (M / 2) > 0, the solution is 2

The maximum value of y = sinx2 + SiNx is______ The minimum value is______ ．

Solution 1: y = 2 + SiNx − 22 + SiNx = 1-22 + SiNx. When SiNx = - 1, we get Ymin = - 1, when SiNx = 1, we get ymax = 13. Solution 2: the original formula {SiNx = 2y1 − y (∵ y ≠ 1)} | 2y1 − y | ≤ 1 {- 1 ≤ y ≤ 13. Ymax = 13, Ymin = - 1. Answer: 13; - 1