If the solution set of a2x-5 is

If the solution set of a2x-5 is

a-5
(a-2)x>-5
A-2 is negative, so divide both sides by A-2, and the inequality sign changes direction
So x

How to find the value range of X & # 178; - 2x-15 > 0
How to write the format? How to ask

A:
Factorization is solvable
x²-2x-15>0
(x-5)(x+3)>0
The solution is: X5

If x + Y > = 0, X & # 178; + Y & # 178;

Draw the feasible region
When y = - 2x + Z is tangent to the feasible region, the intercept Z has a maximum value and a minimum value, which are √ 5 & nbsp; - √ 2 / 2, respectively
Value range [- √ 2 / 2, √ 5]

Let a = {x logx (5x & sup2; - 8x + 3) > 2}, B = {X & sup2; - 2x-a4 power + 1 ≥ 0}, we know that a is really included in B, and find the value range of A
1) Such as the title
2) The solution set of (4) 2x power - (2) 2x + 2x power + 3 < 0
3) Solving inequality: under the root sign (2x + 5) > x + 1
4) Solving inequality: log (x-3) (X & sup2; - 3x-4) < 2 ((x-3) is the base)
5) If the inequality X & sup2; - logmx < 0 holds in the range of (0,1 / 2), then the value range of real number m (M is the base)
6) The solution set of inequality log3 (3-x) ≥ 1 is? (3 is the base)
You can solve any problem you can,

Let a = {x logx (5x ^ 2-8x + 3) ＞ 2}, B = {x ^ 2-2x-a ^ 4 + 1 ≥ 0}, and it is known that a is really included in B, then the value range of a is obtained
logx(5x^2-8x+3)＞logx(x^2)
logx[(5x^2-8x+3)/x^2]>logx(1)
[(5x^2-8x+3)/x^2]>1
(5x^2-8x+3-x^2)/x^2>0
log3 (3-x)