Known function f (x) = x ^ 3 - (k ^ 2-k + 1) x ^ 2 + 5x-2, G (x) = k ^ 2x ^ 2 + KX + 1 Let P (x) = f (x) + G (x). If P (x) is not monotone in the interval (0,3), the value range of K is obtained

Known function f (x) = x ^ 3 - (k ^ 2-k + 1) x ^ 2 + 5x-2, G (x) = k ^ 2x ^ 2 + KX + 1 Let P (x) = f (x) + G (x). If P (x) is not monotone in the interval (0,3), the value range of K is obtained

If P (x) = x + (k-1) x + (K + 5) X-1, p '(x) = 3x + 2 (k-1) x + (K + 5) P (x) is not monotone in the interval (0.3), then p' (x) = 0 has a solution in (0,3). If there is only one solution, then p (0) * P (3) - 57K + 26 > 0, k > - 26 / 7 (k-1) - 3 (K + 5) > 0, k-5k-14 > 0, K70
Given that vectors a and B are two mutually perpendicular unit vectors, | vector C | = 13, vector c * vector a = 3, vector c * vector b = 4, then for any real number T1 and T2
I wish you all the best in the year of the ox!
If and only if T1 = 3, T2 = 4, take the equal sign. The minimum value of | c-t1a-t2b | is 12
The constant term of (R + 2) x square + 3 (R-2) x + R square - 3 = 0 is 1, then r =?
The constant term of the quadratic equation (R + 2) x square + 3 (R-2) x + R square - 3 = 0 is 1
The solution is r = 2 or - 2
Because the coefficient of quadratic term of quadratic equation of one variable is not 0, that is, R + 2 is not equal to 0, so r is not equal to - 2
The solution is r = 2
R & # 178; - 3 = 1 and quadratic coefficient R + 2 ≠ 0
So r = 2
How to calculate 16.5-3x multiplied by 2.5 = x?
16.5-3x times 2.5 = x
16.5-3*2.5 x = x
16.5=8.5 x
x= 16.5/8.5 =1.9411.
Is there a problem with the title? Or is there a bracket?
（16.5-3x） * 2.5 = x
If the limit of (n & # 178; + 1) / (n + 1) - an-b is 0, then a + 2B=
First of all, we have to share
If B = - 1, then the limit is zero
So the above formula is - 1
Given the function f (x) = - x ^ 2 + KX + 5x + 1, G (x) = - LNX + KX, where k ∈ R (1) when k = 1, find the extremum of the number of rows f (x), (2) if the equation f (x) = 0 on X
Let Q (x) = f (x) (x ≤ 0) Q (x) = g (x) (x > 0), is there a solution in the interval (1,2), so that for any point of Q (x) (abscissa is not 0), another unique point can always be found, so that the slopes of tangents at these two points are equal? If there is a solution, there is no reason
1 K = 1, f = - x ^ 2 + 6x + 1, the axis of symmetry is x = 3, so f (3) = - 9-18 + 1 = - 26 is the maximum
2 F = 0 = - x ^ 2 + (K + 5) x + 1 = x ^ 2 - (K + 5) X-1, so the product of the two is - 1, and one root is on (1,2), which is the larger root (K + 5) / 2 + root ({(K + 5) / 2} ^ 2 + 1), so 1
Given three points a (radical 3 + 1,1), B (1,1), C (1,2), then=
Method 1
It is obvious that the vector CA = (√ 3, - 1) and the vector CB = (0, - 1)
Vector Ca · vector CB = 0 + 1 = 1
Vector CA = √ (3 + 1) = 2, vector CB = √ (0 + 1) = 1
The results show that cos ＜ Ca, CB ＞ = vector Ca · vector CB / (︱ vector Ca ︱ vector CB) = 1 / (2 × 1) = 1 / 2,
∴＜CA,CB＞＝60°.
Method 2
｜AB｜＝√［（√3＋1－1）^2＋（1－1）^2］＝√3,
｜AC｜＝√［（√3＋1－1）^2＋（1－2）^2］＝2,
｜BC｜＝√［（1－1）^2＋（1－2）^2＝1.
However, AC ︱ BC ︱ 2 ︱ AC ︱ BC ︱ 2 ︱ AC ︱ BC ︱ 2 ︱ AC ︱ BC ︱ 2 ︱ AC ︱ BC ︱ 2 ︱ BC ︱ AC ︱ BC ︱ 2,
∴＜CA,CB＞＝60°.
In the plane rectangular coordinate system, point C has three points: a (- 1, - 1), B (radical 3-1,0) and C (radical 3-1,2).
1. The module of vector AC and the size of AC.
2. The angle between vector AC and ab
It is known that the square of the quadratic equation x of one variable with respect to X - (M + 2) x + 2m = 0 (M is a constant)
(1) It is proved that the equation always has two real roots
△=b^2-4ac
=(m+2)^2-4*2m
=m^2+4m+4-8m
=m^2-2m+4
=(m-2)^2≥0
So the equation always has two real roots
Discriminant = (M + 2) ^ 2-4 * 2m = m ^ 2 + 4m + 4-8M = m ^ 2-4m + 4 = (m-2) ^ 2
So there are two solutions (the two solutions may be equal)
Δ=(m+2)^2-4*(2m)
=m^2+4m-8m+4
=m^2-4m+4
=(m-2)^2
>=0
So the equation always has two real roots
^2 is the square
Calculate (- 3x ^ 2 + 5) (- 3x ^ 2-5) - x ^ 2 (3x + 4) (3x-4) - 16 (- x) ^ 2
Original formula = (9x ^ 4-25) - x ^ 2 (9x ^ 2-16) - 16x ^ 2
=9x^4-25-9x^4+16x^2-16x^2
=0
If the algebraic formula A ^ 2B ^ (n-1) / 3 and - 3 / 7-a ^ MB & # 178; can be merged, then m ^ n =?
A ^ 2B ^ (n-1) / 3 and - 3 / 7-a ^ MB & # 178; can be merged
It shows that there is the same index
Then M = 2
n-1=2
M = 2, n = 3
So m ^ n = 2 ^ 3 = 8