It is known that the equation x2 + X + P = 0 has two roots x1, X2, P ∈ R. if | x1 | + | x2 | = 3, then the value set of real number P is,

It is known that the equation x2 + X + P = 0 has two roots x1, X2, P ∈ R. if | x1 | + | x2 | = 3, then the value set of real number P is,

The discriminant is greater than 0, so P

It is known that the function f (x) is a linear function, and for any t belonging to R, there is always 3f (T + 1) - 2F (t-1) = 2T + 17, so the expression of F (x) can be obtained

The general formula of a function is y = ax + B, that is, f (x) = ax + B. in the problem, 3f (T + 1) - 2F (t-1) = 2T + 17 is known. Substituting the general formula into 3 [a (T + 1) + b] - 2 [a (t-1) + b] = 2T + 17, the expansion is 3At + 3A + 3b-2at + 2a-2b = 2T + 17at + 5A + B = 2T + 17, that is, a = 25A + B = 17, B = 7 is substituted into the general formula to get f (x) = 2x + 7

It is known that | a | = 2 | B | ≠ 0, and the equation x2 + | a | x + a · B = 0 has a real solution, then the range of the angle between a and B is

Let the angle between a and B be α, then the equation becomes: x ^ 2 + | a | x + | a | B | cos α = 0,
Given that | a | = 2 | B | ≠ 0, the equation becomes: x ^ 2 + 2 | B | x + 2 | B | ^ 2cos α = 0
If the equation is known to have real solution, the discriminant = 4|b | ^ 2-8|b | ^ 2cos α ≥ 0,
The results show that cos α ≤ 1 / 2, π / 3 ≤ α ≤ π,
The angle between a and B is [π / 3, π]

If we know that f (x) is a function of degree and satisfies 3f (x + 1) = 2x + 17, then f (x) = ()
A. 23x+5B. 23x+1C. 2x-3D. 2x+5

Let f (x) = KX + B ∵ 3f (x + 1) = 2x + 17, ∵ 3 [K (x + 1) + b] = 2x + 17, i.e., 3kx + 3K + 3B = 2x + 17 ∵ 3K = 23K + 3B = 17, then we can solve the equation, k = 23, B = 5 ∵ f (x) = 23x + 5, so we choose a

Given the solution x1, X2 of the equation x & sup2; + (A-3) x + 3 = 0 in the range of real number, and X1 > 1, X2 > 1, find the value range of A

Firstly, we guarantee that the equation has roots, that is, Δ > 0 leads to (A-3) & sup2; - 4 * 1 * 3 > 0 leads to solutions x1, X2 of a ＜ 3-2 √ 3 or a ＞ 3 + 2 √ 3 in the range of real numbers, and the conditions of x1 > 1, X2 > 1 can be equivalent transformed into: (x1-1) + (x-1) > 0 (x1-1) * (x2-1) > 0, that is, X1 + x2 > 2

If we know that f (x) is a function of degree and satisfies 3f (x + 1) = 2x + 17, then f (x) = ()
A. 23x+5B. 23x+1C. 2x-3D. 2x+5

Let f (x) = KX + B ∵ 3f (x + 1) = 2x + 17, ∵ 3 [K (x + 1) + b] = 2x + 17, i.e., 3kx + 3K + 3B = 2x + 17 ∵ 3K = 23K + 3B = 17, then we can solve the equation, k = 23, B = 5 ∵ f (x) = 23x + 5, so we choose a

How many solutions does the equation (x2 + 3x) 2 + 9 (x2 + 3x) - 22 = 0 have in the real range

Let t = x2 + 3x change the original equation to T & # 178; + 9t-22 = 0, i.e. (T + 11) (T-2) = 0, i.e. t = - 11 or T = 2. When t = - 11, X & # 178; + 3x = - 11, i.e. X & # 178; + 3x + 11 = 0, because Δ = 3 & # 178; - 4 * 1 * 11 < 0, i.e. the equation has no solution. When t = 2, X & # 178; + 3x = 2, i.e. X & # 178; + 3x-2 = 0, i.e. Δ = 3 & # 178

If f (x + 1) and f (x-1) are both odd functions, then ()
A. F (x) is an even function B. f (x) is an odd function C. f (x) = f (x + 2) d. f (x + 3) is an odd function

∵ f (x + 1) and f (x-1) are both odd functions. The function f (x) is symmetric with respect to point (1,0) and point (- 1,0), f (x) + F (2-x) = 0, f (x) + F (- 2-x) = 0, so f (2-x) = f (- 2-x). The function f (x) is a periodic function with period T = [2 - (- 2)] = 4

If X1 and X2 are two real number solutions of the equation 2 ^ x = 2 ^ (1 / x) + 1, find X1 + x2

Equation: 2 ^ x = 2 ^ (1 / x) + 1
There is only one real solution

Why is the set of all real roots of equation x = x {0,1}

Solving the equation x ^ 2 = x
Deformation x ^ 2-x = 0
That is, X (x-1) = 0
We get x = 0 or x = 1
That is to say, all solutions of x ^ 2 = x are 0 and 1
So the set of its real roots is {0,1}
Hope to help you.