Then, if there is no system of inequalities, the range of solution is XA-1————

Then, if there is no system of inequalities, the range of solution is XA-1————

a-1≥2a+1
So a ≤ - 2
A is less than - 2
In the number axis method, if point 2A + 1 is on the right side of point A-1, there is a solution; if point 2A + 1 is on the left side of point A-1, there is no solution, that is 2A + 1
If the system of inequalities x < A + 1 x > 2a-1 has no solution, then the value range of a is -. If there is a solution, then a------
Similarly, if the system of inequalities x < A + 1 x > 2a-1 has no solution, then the value range of a is a ≥ 2. If there is a solution, then a2a-1,
A
It is known that the solution set of equation 2 (x-a) = x-a + 1 is suitable for inequality x ≥ 1, then the value range of a is---------
The solution consists of 2 (x-a) = x-a + 1
We get 2x-2a = x-a + 1
That is, x = a + 1
The solution set of equation 2 (x-a) = x-a + 1 of X is suitable for inequality x ≥ 1
That is x ≥ 1
That is, a + 1 ≥ 1
That is, a ≥ 0
If the zero point of function f (x) = ax + B is 2, then the zero point of function g (x) = bx2 ax is ()
A. 0，2B. 0，12C. 0，-12D. 2，12
∵ function f (x) = ax + B has a zero point of 2, ∵ 2A + B = 0, {B = - 2A, ∵ g (x) = bx2 AX = - 2ax2 AX = - ax (2x + 1), ∵ ax (2x + 1) = 0 {x = 0, x = - 12 ∵ function g (x) = bx2 ax has zero points of 0, - 12
Given the function f (x) = 1 / 2x-3x-3 / 4, find the value range of X that makes the function value greater than 0
1. (x-6x + 9-9-9) - 3 / 4 = 1 / 2 (x-6x-6x + 9-9-9) - 3 / 4 (6-6x-6x-9-9-9) - 3 / 4 = 1 / 2 (x-6-6x-9-9-9-9-9-9) - 3 / 4 (4-2 (x-3) - 21 / 2 (x-6x-9-9-9-9-9) - 3 / 4 = 1 / 2 (X-2 (x-3) - 21 / 2 (x-3) - 21 / 2 (x-3) - 21 / 2 (x-3) - 21 (2 / 2) (21 / 2) (21 / 2 (X-2 / 2, 4x-20x-25, 4x-204x-204x-204x-20x-20x-4x-20x-20x-20x-25-25 (4x4x4x-20x-20x-20x-25-25-25-25-25-25 2 ＜ x ＜ (5 + 5 √ 2) / 2 3, - 3x + 5x-4 ＞ 0 Since the discriminant △ = (- 5) - 4 × 3 × 4 < 0 ｜ 3x-5x + 4 < 0 holds, the solution set of ｜ 3x + 5x-4 > 0 is x ∈ R
If the zero point of function f (x) = ax + B is 2, then the zero point of function g (x) = bx2 ax is ()
A. 0，2B. 0，12C. 0，-12D. 2，12
∵ function f (x) = ax + B has a zero point of 2, ∵ 2A + B = 0, {B = - 2A, ∵ g (x) = bx2 AX = - 2ax2 AX = - ax (2x + 1), ∵ ax (2x + 1) = 0 {x = 0, x = - 12 ∵ function g (x) = bx2 ax has zero points of 0, - 12
Let f (x) = {2 ^ XX2, then the value range of a is
Let f (x) = {2 ^ x 2, then the value range of a is?
When A2
A>1
∴1log3(3^2)
2a+1>9
2a>8
A>4
∴a∈(1,2)∪(4,+∞)
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2 ^ a > 2 (A2 (a > = 2), the solution is a ∈ (1,2) ∪ (4, + ∞)
The most known zeros of the function Λ (x + 2) = x (x) = 0,2
The small value is - 1 / 8 (1) to find the analytic expression of function f (x)
F (0) = C = 0, f (1) = a + B = 0, f (- B / 2a) = B ^ 2 / 4a-b ^ 2 / 2A = B ^ 2 / 4A = - 1 / 8, that is, a = 2B ^ 2, then 2B ^ 2 + B = 0, that is, B = 0 or - 1 / 2, excluding B = 0, that is, B = - 1 / 2, a = 1 / 2, so f (x) = 1 / 2x ^ 2-1 / 2x
Let f (x) be written as f (x) = a (x-1) and answer: F (x) = ax (x-1). When x = 0.5, the minimum value can be obtained. In this way, a = 0.5 can be solved, so f (x) can be solved.
If f (3x-1) = x & # 178; + 2x + 4, then f (x) =?
Let 3x-1 = t, then x = (T + 1) / 3, then
f(t)=[(t+1)/3]²+2*(t+1)/3+4
=(t²+2t+1)/9+(2t+2)/3+4
=(t²+2t+1+6t+6+36)/9
=(t²+8t+43)/9
therefore
f(x)=(x²+8x+43)/9
If the two zeros of function f (x) = x ^ 2-ax-b are 2 and 3, find the zeros of function g (x) = BX ^ 2-ax-a
Such as the title
The two zeros of F (x) = x ^ 2-ax-b are 2 and 3
According to the relationship between root and coefficient
a=2+3=5
b=2*3=6
g(x)=bx^2-ax-a
=6x^2-5x-5=0
X1 = (5 + radical 145) / 12
X2 = (5-radical 145) / 12
That is two zeros