Let f (x) be an odd function defined on R, and for any real number x, f (x + 2) = - f (x). When x ∈ [- 2,0], f (x) = 2x-x ^ 3 (1) Prove that f (x) is a periodic function (2) When x ∈ [2,4], find the analytic expression of F (x) (3) . find the value of F (0) + F (1) + F (2) +... + F (2011) If x ∈ [0,2], f (x) = 2x-x ^ 3

Let f (x) be an odd function defined on R, and for any real number x, f (x + 2) = - f (x). When x ∈ [- 2,0], f (x) = 2x-x ^ 3 (1) Prove that f (x) is a periodic function (2) When x ∈ [2,4], find the analytic expression of F (x) (3) . find the value of F (0) + F (1) + F (2) +... + F (2011) If x ∈ [0,2], f (x) = 2x-x ^ 3

Because f (x + 2) = - f (x), f (x + 4) = - f (x + 2)
So f (x) = f (x + 4)
Then the period of F (x) is 4
When x ∈ [- 2,0], - x ∈ [0,2],
Then f (- x) = 2 (- x) - (- x) ^ 2 = - 2x - x ^ 2,
Because f (x) is an odd function,
So f (x) = - f (- x) = - [- 2x - x ^ 2] = 2x + x ^ 2 (when x ∈ [- 2,0])
When x ∈ [2,4], x-4 ∈ [- 2,0],
So f (x-4) = 2 (x-4) + (x-4) ^ 2
Because the period of F (x) is 4,
So f (x) = f (x-4) = 2 (x-4) + (x-4) ^ 2
=X ^ 2-6x + 8 (when x ∈ [2,4])
When x ∈ [0,2], f (x) = 2x-x ^ 2
When x ∈ [2,4], f (x) = = x ^ 2-6x + 8
So f (0) = 0, f (1) = 1, f (2) = 0, f (3) = - 1
f(0)+f(1)+f(2)+f(3)=0.
Because the period of F (x) is 4,
So f (0) + F (1) + F (2) + +f(2012)
= [f(0)+f(1)+f(2)+f(3)]+[ f(4)+f(5)+f(6)+f(7)]+…… +[ f(2008)+f(2009)+f(2010)+f(2011)]+ f(2012)
=0+0+…… +0+ f(2012)
= f（0）
=0.
⑴f(x+2)=-f(x)
f(x+2)=-f(x+4)
f(x)=f(x+4) T=4
⑵x∈[-2,0]
x+4∈[2,4]
f(x)=f(x+4)
∴f(x)=2x-x^3
⑶f(0)=0
f(1)=1
So you can get the period. Follow up: just hit the wrong question
Two polynomials AB, B is the square of 4x minus the square of 5x minus 6, find a + B, if a + B as A-B, the result is the square of - 7x + 10x + 12, a + B
Two polynomials AB, B is the square of 4x minus the square of 5x minus 6..., find a + B, if a + B is regarded as A-B, the result is the square of - 7x + 10x + 12, what is the correct answer of a + B?
A-B, the result is - 7x squared + 10x + 12,
A = - 7x square + 10x + 12 + B = - 7x ^ 2 + 10x + 12 + 4x ^ 2-5x ^ 2-6 = - 8x ^ 2 + 10x + 6
A+B==-8x^2+10x+6+4x^2-5x^2-6=-9x^2+10x
To solve the equation, we need to test 5x-4 * 9 = 24
5x-4*9=24
5x-36=24
5x=24+36
5x=60
x=60÷5
x=12
Test: left side of equation = 5 × 12-4 × 9 = 60-36 = 24 = right side of equation
5x-4*9=24
5x-36=24
5x=60
x=12
Test: substitute x = 12 into the equation
Left = 5 × 12-4 × 9 = 60-36 = 24 = right
So x = 12 is the solution of the equation
Twelve
5x-4×9=24
5x-36=24
5x=24+36
5x=60
x=12
test
Left = 5 × 12-4 × 9
=60-36
=24
Right = 24
It is known that the equation 2x & # 178; + MX-1 = 0 about X has two real roots - 2, N. find the value of M, n
Sum of quadratic equations of one variable = - M / 2 = - 2 + n
Two product = - 1 / 2 = - 2n
To solve this system of linear equations with respect to m and N, we obtain
m=7/2 n=1/4
There's a mistake upstairs, M
The M / 2 = - 2 + n with negative sum of two is obtained by using Weida's theorem
The product of two - 2n = - 1 / 2, so n = 1 / 4, M = negative 7 / 2
If the function f (x) = 2x & # 178; - MX + 3 is a decreasing function in the interval (- ∞, - 2], then the value range of real number m is?
The symmetry axis of F (x) = 2x & # 178; - MX + 3 is x = m / 4,
Then M / 4 ≥ - 2,
That is, m ≥ - 8
In finding the sum of a polynomial and the square of 4x minus 5x and then minus 6, Xiao Hu mistakenly regarded "sum" as "difference", making the calculation result - 7x + 10x + 12
Find out the correct calculation result
Note: the calculation result here is - 7x + 10x + 12, which should be - 7x & # 178; + 10x + 12! From the meaning of the title, we can see that the unknown polynomial can be expressed as: (- 7x & # 178; + 10x + 12) + (4x & # 178; - 5x-6) = - 3x & # 178; + 5x + 6, so the correct calculation result should be: - 3x & # 178; + 5x + 6 + (4x & # 178; - 5x -
The unknown polynomial can be expressed as:
(-7x² +10x+12)+(4x² -5x-6)=-3x² +5x+6
Therefore, the correct calculation results should be as follows:
-3x² +5x+6+(4x² -5x-6)＝x²
Note: the calculation result here is - 7x + 10x + 12, which should be - 7x & # 178; + 10x + 12!
Analysis:
The unknown polynomial can be expressed as:
(-7x² +10x+12)+(4x² -5x-6)=-3x² +5x+6
Therefore, the correct calculation results should be as follows:
-3x² +5x+6+(4x² -5x-6)＝x²
24-0.8 (x-2.5) = 0.5x, how to solve the equation?
24-0.8(x-2.5)=0.5x
24-0.8x+2=0.5x
24+2=0.8x+0.5x
26=1.3x
x=20
Multiply the number in brackets and move it. thank you
24-0.8X+2=0.5X
-1.3X = -2-24
X= 20
Just be careful. It's going to change in the middle
If the two roots of the equation 2x2 + mx-n = 0 of X are - 1 and 3, then the factorization result of 2x2 + mx-n is ()
A. （x+1）（x-3）B. 2（x+1）（x-3）C. （x-1）（x+3）D. 2（x-1）（x+3）
∵ the two roots of the equation 2x2 + mx-n = 0 are - 1 and 3, ∵ 1 + 3 = - m2, - 1 × 3 = - N2, ∵ M = - 4, n = 6. ∵ 2x2-4x-6 = 2 (x2-2x-3) = 2 (x + 1) (x-3)
If the function f (x) = x2 + 2 (A-1) x + 2 is a decreasing function on (- ∞, 4], then the value range of real number a is ()
A. a≤-3B. a≥-3C. a≤5D. a≥5
∵ f (x) = x2 + 2 (A-1) x + 2 = (x + A-1) 2 + 2 - (A-1) 2, its axis of symmetry is: x = 1-A ∵ function f (x) = x2 + 2 (A-1) x + 2 is a decreasing function on (- ∞, 4], so a is selected
When Xiao Ming did a math problem, he mistakenly regarded "a + B" as the result of "A-B", and the answer was - 7x square + 10x + 12. He knew that B = 4x square - 5x-6,
a-b=a-(4x^2-5x-6)=a-4x^2+5x+6=-7x^2+10x+12
So a = - 3x ^ 2 + 5x + 6
a+b=(-3x^2+5x+6)+(4x^2+5x+6)=x^2
What do you want?