If x squared minus ax minus 15 equals [x plus 1] times [x minus 15], then a equals This is bracket

If x squared minus ax minus 15 equals [x plus 1] times [x minus 15], then a equals This is bracket

analysis
x^2-Ax-15=(x+1)(x-15)
=x^2+x-15x-15
=x^2-14x-15
So a = 14
A {the square of x minus 8x plus 15 equals 0} and B {ax minus 1 equals 0} if B is contained in a, then the number of subsets of a
A = {3,5} and B is a subset of A
B=｛X=1/a｝
1) A = 0 no
2) A is not equal to 0,4
It should be. It's not bad for me to be wrong
A = 1 / 5 or 1 / 3 or 0
The number of subsets that make up a set is 2 ^ 3 = 8
If the operation a ⊙ B = B, a ≥ B or a ⊙ B = a, a is defined
F (x) = x ⊙ (2-x) when x ≥ 2-x, that is, X ≥ 1, f (x) = 2-x ≤ 1; when x < 2-x, that is, x < 1, f (x) = x < 1
Therefore, the range is f (x) ≤ 1
It is known that the solution set of inequality | X-1 | ax about X is m. if M ∩ z = {1,2}, then the value range of real number a is---
It is known that the integer solutions satisfying the inequality are only 1 and 2,
So we can get 0
Define the operation a * b = {a (a is less than or equal to b), B (a is greater than B). Then what is the range of F (x) = 1 * 2
Sorry, the last question to be changed is: what is the range of the x power of the function f (x) = 1 * 2
By definition, a * B means the smaller of a and B, that is, min {a, B}, 1 * 2 = 1
F (x) = 1 * 2 = 1 is a constant function with a range of {1}
If it's (1 * 2) ^ x, the result doesn't change
If it is 1 * (2 ^ x), when x0, when x > 0, f (x) = 1, range (0,1)]
It is known that the solution of the quadratic inequality ax ^ 2 + BX + C ＞ - 2x is 1 < x < 3
(1) If the equation AX ^ 2 + BX + C + 6A = 0 has two unequal real roots, find the analytic expression of quadratic function
(2) The maximum value of function y = ax ^ 2 + BX + C is a positive number, and the value range of real number a is calculated
(1) The quadratic inequality ax ^ 2 + BX + C ＞ - 2x can be reduced to AX ^ 2 + (B + 2) x + C ＞ 0
Because its solution is 1 < x < 3, a < 0, and the original inequality is equivalent to a (x-1) (x-3) > 0
We can get - 4A = B + 2, that is, B = - 2-4a
3a=c c=3a
Then the equation AX ^ 2 + BX + C + 6A = 0 can be reduced to AX ^ 2 - (2 + 4a) x + 9A = 0
Because it has two equal real roots
So a < 0
Δ=(2+4a)^2-36a^2=0
The solution is a = - 1 / 5 or 1 (rounding off)
A = - 1 / 5
Then the analytic expression of quadratic function is y = - 1 / 5x ^ 2-6 / 5x-3 / 5
(2)y=ax^2-(2+4a)x+3a
The maximum value is a positive number
That is [4A * 3A - (2 + 4a) ^ 2] / (4a) > 0
a＜0
The root 3-2 ＜ a ＜ 0 or a ＜ - root 3-2 is obtained
There should be some problems in the first question, I changed it. The analytic expression of quadratic function is also solved according to the second question
The general idea is this, if there are miscalculations, please include
a0,a!=0
c-b^2/4a>0
The solution of the quadratic inequality ax ^ 2 + BX + C ＞ - 2x about X is 1 < x < 3. knowable
The solution of ax ^ 2 + BX + C + 2x = 0 is 1.3
So a + B + C + 2 = 0
9a+3b+c+6=0
-B / 2A = 2 Veda theorem
A, B, C are solved
Other solvable
Function image is very important
Definition operation: a ⊗ B = a (a ≤ b) B (a > b), then the range of function f (x) = 2x ⊗ 2-x is______
When 2x ≤ 2-x, i.e. x ≤ 0, the function f (x) = 2x ⊗ 2-x = 2x; when 2x ⊗ 2-x, i.e. x > 0, the function f (x) = 2x ⊗ 2-x = 2-x  f (x) = 2x, X ≤ 0 (12) x, x > 0. From the figure, the range of function f (x) = 2x ⊗ 2-x is: (0,1]. Therefore, the answer is: (0,1]
It is known that the Real Coefficient Quadratic Inequality ax ^ 2 + BX + C > = 0 (a = 0 (a) for X
One variable quadratic inequality ax & # 178; + BX + C ≥ 0 (A0 (parabola y = ax & # 178; + BX + C opening must face up, otherwise the original inequality cannot be less than 0 on the whole R.)
C ≥ 0 (by substituting x = 0 into the original inequality)
Let B = B / A, C = C / A, then
B> 1 (from 0
On the function f (x) = n (x-1) &# 178; + 2m-3 (n ≠ 0) of X, the domain of definition and the range of value are [1, M] where m > 1, find the analytic expression of F (x)
On the function f (x) = n (x-1) &# 178; + 2m-3 (n ≠ 0) of X, the domain of definition and the range of value are [1, M] where m > 1, find the analytic expression of F (x)
Analysis: the function f (x) = n (x-1) &# 178; + 2m-3 (n ≠ 0) of ∵ X has the domain of [1, M] where m > 1
F’(X)=2Nx-2N=0==>x=1
The axis of symmetry of the function is x = 1
The extreme value is 2m-3
When n > 0, the function has a minimum, 2m-3 = 1 = > m = 2,
N(M-1)^2+1=M==>N=1
∴f(x)=(x-1)^2+1
When nm = 3
N(3-1)^2+3=1==>M=-1/2
∴f(x)=-1/2(x-1)^2+3
When n > 0, f (x) = 1 / (m-1) * (x-1) 2 + 1
When n
The solution set of inequality ax whose square + BX + C is less than 0 (a is not equal to 0) is r, then ()
A. A less than 0 △ less than 0 B. a less than 0 [
A
When a is less than zero, the opening is downward, and when △ is less than 0, there is no intersection with X axis