Given ax ^ 4 + BX ^ 3 + CX ^ 2 + DX + e = (X-2) ^ 4, find the value of a + B + C + D + E First question: Known ax ^ 4 + BX ^ 3 + CX ^ 2 + DX + e = (X-2) ^ 4 Find the value of a + B + C + D + e Try to find the value of a + C Second question: If the value of polynomial (2mx ^ 2-y ^ 2 + 3x + 1) - (5x ^ 2-5y ^ 2 + 3x) is independent of X, Find the value of 4m ^ 2 - (4m-5) + 6m

Given ax ^ 4 + BX ^ 3 + CX ^ 2 + DX + e = (X-2) ^ 4, find the value of a + B + C + D + E First question: Known ax ^ 4 + BX ^ 3 + CX ^ 2 + DX + e = (X-2) ^ 4 Find the value of a + B + C + D + e Try to find the value of a + C Second question: If the value of polynomial (2mx ^ 2-y ^ 2 + 3x + 1) - (5x ^ 2-5y ^ 2 + 3x) is independent of X, Find the value of 4m ^ 2 - (4m-5) + 6m

solution
Let x = 1, then
(1-2)^4=a+b+c+d+e
∴a+b+c+d+e=(-1)^4=1 ①
Let x = - 1, then
(-1-2)^4=a-b+c-d+e
∴a-b+c-d+e=(-3)^4=81 ②
① + 2
2a+2c+2e=82
∴a+c+e=41
Let x = 0, then
(0-2)^4=e
∴e=16
∴a+c=41-16=25
The value of (2mx ^ 2-y ^ 2 + 3x + 1) - (5x ^ 2-5y ^ 2 + 3x) has nothing to do with X,
=(2mx²-5x²)+(3x-3x)+(-y²+5y²)+1
=(2m-5)x²+4y²+1
∴2m-5=0
∴m=5/2
Find the value of 4m ^ 2 - (4m-5) + 6m
=4m²-4m+5+6m
=4m²+2m+5
=4×(25/4)+2×(5/2)+5
=25+5+5
=35

Given (x + 1) 5 = ax5 + bx4 + cx3 + DX2 + ex + F, find the value of the following formula: (1) a + B + C + D + e + F; (2) B + C + D + E; (3) a + C + E

（1）（x+1）5，=（x+1）2×（x+1）2×（x+1），=（x2+2x+1）（x2+2x+1）（x+1），=（x4+4x3+6x2+4x+1）（x+1），=x5+5x4+10x3+10x2+5x+1，∵（x+1）5，=ax5+bx4+cx3+dx2+ex+f，∴a=1，b=5，c=10，d=10，e=5，f=1，...

If f (x) = MX ^ 2 - (M-4) x + 1 has at least one zero on the left side of the origin, find the value range of real number M
Don't copy the answer directly on the Internet. My method is to find the value range of m on the right side first, and then on the contrary, the value range of m on the left side is the positive solution ~ but I can find the value range of m on the right side at last, M > 0, M > 4, m ≤ 6-2 √ 5, m ≥ 6 + 2 √ 5, and then I won't calculate. How should I take the value? The answer is (negative infinity, 6-2 √ 5]

If M = 0, then f (x) is a function of degree, that is, f (x) = 4x + 1, when f (x) = 0 (zero point)
Then x = - 1 / 4, which is consistent with the title (there is at least one zero on the left side of the origin, that is, x < 0), so m = 0
If M ≠ 0, then f (x) is a quadratic function. If there is at least one zero on the left side of the origin, then △≥ 0
That is, m ≤ 6-2 √ 5, m ≥ 6 + 2 √ 5
To sum up, m belongs to negative infinity, 6-2 √ 5]
LZ algorithm do not understand, a bit confused

Let f (x) be an even function on R, increasing on the interval (- ∞, 0), and f (2a square + A + 1) < f (2a square - 2A + 3). Find the value range of A

2a^2+a+1=2(a+1/4)^2+7/8>0 2a^2-2a+3=2(a-1/2)^2+5/2>0
The function f (x) is an even function on R, if it increases on the interval (- ∞, 0), then it decreases on (0, + ∞), f (2a ^ 2 + A + 1) < f (2a ^ 2-2a + 3)
So 2A ^ 2 + A + 1 > 2A ^ 2-2a + 3 a > 2 / 3

Find the extreme value of F (x) (2) a when f (x) = (2-A) LNX + 1 / x + 2aX (a ≤ 0) (1) a = 0

f'(x)=(2-a)/x-1/x²+2a
When a = 0, f '(x) = 2 / X-1 / X & # 178; = 0, the solution is x = 1 / 2, f (1 / 2) = 2-4ln2
(2) A0 means that f increases; when 1 / X ∈ (0,2] u [- A, + ∞), f '(x) < 0, that is, f decreases
That is, when x ∈ (- 1 / A, 1 / 2), f (x) increases monotonically; when x ∈ (0, - 1 / a] u [1 / 2, + ∞), f (x) decreases monotonically
When a = - 2, f '(x) ≤ 0, f (x) decreases monotonically on (0, + ∞)
When a ∈ (- 2,0), when 1 / X ∈ (- A, 2), f '(x) > 0, that is, f increases; when 1 / X ∈ (0, - A] u [2, + ∞), f' (x) < 0, that is, f decreases
That is, when x ∈ (1 / 2, - 1 / a), f (x) increases monotonically; when x ∈ (0,2] u [- 1 / A, + ∞), f (x) decreases monotonically
(3) Check the title, there is a problem

Given the function f (x) = ax − ax − 2lnx (a ≥ 0), if f (x) is a monotone function in its domain of definition, the value range of a is obtained

The original function definition domain is (0, + ∞) ｜ f ′ (x) = a + AX2 − 2x = AX2 − 2x + AX2 ∵ the function f (x) is a monotone function in the definition domain (0, + ∞), and ｜ f '(x) ≤ 0 or F' (x) ≥ 0 is constant in (0, + ∞) (1) when a = 0, f ′ (x) = − 2x < 0 is constant in (0, + ∞), and ｜ a = 0 satisfies the problem (2) when a > 0, let g (x) = ax2-2x + a (x ∈ (0, + ∞) ）From the meaning of the title, we know that △ = 4-4a2 ≤ 0  a ≤ - 1 or a ≥ 1 and ∵ a > 0  a ≥ 1, so the value range of a is: a = 0 or a ≥ 1

Given the function f (x) = ax − ax − 2lnx (a ≥ 0), if f (x) is a monotone function in its domain of definition, the value range of a is obtained

The definition domain of the original function is (0, + ∞) ｜ f ′ (x) = a + AX2 − 2x = AX2 − 2x + AX2 ∵ the function f (x) is a monotone function in the definition domain (0, + ∞), f '(x) ≤ 0 or F' (x) ≥ 0 is constant in (0, + ∞) (1) when a = 0, f ′ (x) = − 2x < 0 is constant in (0, + ∞), a = 0 is full

If f (x) > = f (3) is a positive integer for any x, then the value range of real number C?

From the fundamental inequality a + B ≥ 2 radical ab
The domain is 0 to positive infinity,
So f (x) ≥ 2 radical C
So f (3) = 2 radical C
3 + C / 3 = 2 radical C
c=9

Let f (x) = px-2lnx. (1) if P > 0, find the minimum value of F (x); (2) if G (x) = f (x) - Px is a monotone function in its domain, find the value range of P

(1) ∵ f ′ (x) = p-2x = PX − 2x, Let f ′ (x) = 0, get x = 2p. ∵ P > 0, as shown in the table above, when x = 2p, f (x) has a minimum of 2-2ln2p. (4 points), and this minimum is also the minimum, so when x = 2p, f (x) has a minimum of 2-2ln2p. (5 points) (2) because g (x) = f (x) - Px = px-px-2lnx, then G ′ (x) = P + px2-2x = PX2 − + PX2, From the fact that the function g (x) = f (x) - Px is a monotone function in its domain of definition, it is obtained that G '(x) ≥ 0 is constant for X ∈ (0, + ∞) or G' (x) ≤ 0 is constant for X ∈ (0, + ∞). ① when p = 0, G '(x) = - 2x ＜ 0 is constant for X ∈ (0, + ∞). At this time, G (x) is a decreasing function in its domain of definition, which meets the requirements. ② when p > 0, G' (x) ≤ 0 is constant for X ∈ (0, + ∞)+ From G ′ (x) ≥ 0 to X ∈ (0, + ∞), we can get px2-2x + P ≥ 0 to X ∈ (0, + ∞), that is, P ≥ 2xx2 + 1 to X ∈ (0, + ∞), when x ∈ (0, + ∞), 2xx2 + 1 = 2x + 1x ≤ 1, ∧ P ≥ 1 (9 points). ③ when p < 0, G ′ (x) ≥ 0 to X ∈ (0, + ∞) is impossible, from G ′ (x) ≤ 0 to X ∈ (0, + ∞) Constant holds that px2-2x + P ≤ 0 holds constant for X ∈ (0, + ∞), that is, P ≤ 2xx2 + 1 holds constant for X ∈ (0, + ∞); ∪ [1, + ∞).. (12 points) when x ∈ (0, + ∞), 2xx2 + 1 ∪ [1, + ∞).. (12 points)

It is known that a = {x | 1 / 2

In the interval [1 / 2,2], G (x) = 2x + 1 / X & # 178; = x + X + 1 / X & # 178; ≥ 3, when x = 1, take the equal sign, that is: the minimum value of G (x) = 2x + 1 / X & # 178; is 3, when x = 1
So: F (x) = x & # 178; + PX + Q is also the minimum value of 3 when x = 1
Because 1 / 2 < 1 < 2
So: (1,3) is the vertex coordinates of F (x) = x & # 178; + PX + Q
Because f (x) = x & # 178; + PX + q = (x + P / 2) &# 178; + Q-P & # 178 / 4
So: - P / 2 = 1, Q-P & # / 4 = 3
So: P = - 2, q = 4
So: F (x) = x & # 178; - 2x + 4 = (x-1) &# 178; + 3
So: when x = 2, it gets the maximum value of 4 in the interval [1 / 2,2]