It is known that the minimum value of quadratic function FX is 1 and F0 = F2 = 3 If the image of y = FX is always above the image of y = 2x + 2m + 1, try to determine the value range of real number M

It is known that the minimum value of quadratic function FX is 1 and F0 = F2 = 3 If the image of y = FX is always above the image of y = 2x + 2m + 1, try to determine the value range of real number M

When y = a (x-1) &# 178; + 1 (a ＞ 0) x = 0, y = a + 1 = 3, a = 2  y = 2 (x-1) &# 178; + 1 = 2x & # 178; - 4x + 3 ᙩ y = FX image is always above the image of y = 2x + 2m + 1, G (x) = 2x & # 178; - 4x + 3 - (2x + 2m + 1) = 2x & # 178
From F (0) = f (2) = 3
So let f (x) = ax (X-2) + 3
If the minimum value of F (x) is 1, a = 2 can be obtained
The image of y = FX is always above the image of y = 2x + 2m + 1
That is f (x) - 2x-2m-1 > 0
2x²-6x-2m+2＞0
From △ = 36 + 8 (2m + 2) > 0
M ＞ - 27 / 8
Given the quadratic function f (x) = ax ^ 2 2x C and the maximum value of F (1-x) = f (1 x) is 4, the analytic expression of this function is obtained
Quadratic function f (x) = ax ^ 2 + 2x + C
∵f（1-x）=f（1+x）
The f (x) image is symmetric with respect to x = 1 [f (0) = f (2)]
∴-2/(2a)=1,a=-1
∴f(x)=-x²+2x+c=-(x-1)²+c+1
The maximum value of F (x) is 4
∴c+1=4,c=3
The analytic expression of the function is y = - X & # 178; + 2x + 3
The maximum value is 4,
The analytic expression of this function is y = - X & # 178; + 2x + 3
It is known that the quadratic function f (x) satisfies f (x + 1) - f (x) = 2x and f (0) = 1
The analytic formula of 1-ball f (x)
2. If x ∈ [- 1,1], f (x) > 2x cannot be a sphere real number M
Since f (x) is a quadratic function, Let f (x) = ax ^ 2 + BX + C, then C = 1F (x) = ax ^ 2 + BX + 1F (x + 1) = a (x + 1) ^ 2 + B (x + 1) + 1 = ax ^ 2 + (B + 2a) x + A + B + 1F (x + 1) - f (x) = {ax ^ 2 + (B + 2a) x + A +
It is known that the quadratic function y = one-half of the square of X + x-five-half,
① The coordinates of the vertex and the axis of symmetry are obtained by the matching method;
② If the intersection points of parabola and X axis are a and B, find the length of line ab
①∵y=½x²+x-(5/2)
=½(x+1)²-3
The vertex coordinates of the quadratic function are (- 1, - 3), and the axis of symmetry is a straight line x = - 1
② Let ﹣ x ﹣ x ﹣ x = 0
The solution is: x2 = - 6;
Then AB = - 1 + √ 6) - (- 1 - √ 6) = 2 √ 6
So, the length of line AB is 2 √ 6
The rational numbers a, B, C are not 0, and a + B + C = 0. Let x = |a | / B + C + |b | / C + A + |c | / A + B ì, try to find the value of X to the 19th power plus 2x + 2004
There is absolute value in absolute value. Can you see it
a+b+c=0
Then - a = B + C - B = a + C - C = a + B
If a, B and C are all greater than 0, then a + B + C > 0
If a, B and C are all less than 0, then a + B + C
Because a + B + C = 0, B + C = - A, so | a | / (B + C) = | a | / (- a), the other two formulas can be obtained in the same way.
Because a + B + C = 0, at least one of the three numbers a, B and C is greater than 0, and one of them is less than 0.
Therefore, in the three formulas of | a | / (B + C), | B | / (c + a), | C | / (a + b), there must be one formula equal to 1 and one formula equal to - 1. The two formulas add up to 0. So x = | (+ 1 or - 1) | = 1.
So you can get... Unfolded
Because a + B + C = 0, B + C = - A, so | a | / (B + C) = | a | / (- a), the other two formulas can be obtained in the same way.
Because a + B + C = 0, at least one of the three numbers a, B and C is greater than 0, and one of them is less than 0.
Therefore, in the three formulas of | a | / (B + C), | B | / (c + a), | C | / (a + b), there must be one formula equal to 1 and one formula equal to - 1. The two formulas add up to 0. So x = | (+ 1 or - 1) | = 1.
So the result is 2007. Put it away
The answer is 2001, just mental arithmetic
X minus 3 / - 5 = 3x + 4 / 15,
(x-3)/-5=(3x+4)/15
-3(x-3)=3x+4
-3x+9=3x-4
-3x-3x=-13
-6x=-13
x=13/6
The title is not clear
Your question is not clear. Is it X - (3 / - 5) = 3x + 4 / 15 or (x-3) / - 5 = 3x + 4 / 15? Follow up: x-3 / 5
Factorization: (x ^ 2Y ^ 2 + 1) ^ 2-4x ^ 2Y ^ 2
I'm sorry. My wealth is gone. Can you help me answer the questions,
(x^2y^2+1)^2-4x^2y^2
=(x^2y^2+1)^2-（2xy)^2
=(x^2y^2+1+2xy)(x^2y^2+2xy+1)
=(xy-1)^2(xy+1)^2
The image of quadratic function of the square of y = negative half
Let X be a rational number, then () (a) 2008x (b) x + 2008 (c) must be positive in the following expressions
Let X be a rational number, then ()
（A）2008x （B）x + 2008 （C）|2008x| （D）|x| + 2008
The title is incomplete
A. B is not right
C. D correct
1 / 2 (x-4) - (3x + 4) = - 15 / 2 is a process, which can be used to solve linear equation with one variable
There should be inspection
1/2(x-4）-（3x+4）=-15/2
Multiply both sides by two at the same time
x-4-2(3x+4)=-15
x-4-6x-8=-15
-5x=-15+12
-5x=-3
x=3/5
1/2(x-4）-（3x+4）=-15/2
x-4-3(3x+4)=-15
x-4-9x-12=-15
8x=-1
X = - 1 / 8 question: that's right for so many