It is known that the function f (x) is an odd function defined on (- ∞, + ∞). If for any real number x ≥ 0, f (x + 2) = f (x), and if x ∈ [0,2], f (x) = log2 (x + 1), then the value of F (- 2011) + F (2012) is () A. -1B. -2C. 2D. 1

It is known that the function f (x) is an odd function defined on (- ∞, + ∞). If for any real number x ≥ 0, f (x + 2) = f (x), and if x ∈ [0,2], f (x) = log2 (x + 1), then the value of F (- 2011) + F (2012) is () A. -1B. -2C. 2D. 1

∵ for any real number x ≥ 0, there is a period of F (x + 2) = f (x) in [0, + ∞), t = 2. The∵ function f (x) is an odd function defined on R, so f (- 2011) + F (2012) = - f (2011) + F (2012) = - f (2011) + F (2012) = - f (1) + F (0) if x ∈ [0
X1 and X2 are the two roots of the equation 2x ^ 2-6x + 3 = 0. The value of 1 / X1 + 1 / X2 is
According to Weida's theorem, we get the following results
x1+x2=-(-6/2)=3 x1*x2=3/2
Then: 1 / X1 + 1 / x2
=（x1+x2)/(x1*x2)
=3/3/2
=2 .
Using Weida theorem
2x^2-6x+3=0
x1+x2=3
x1x2=3/2
1/x1+1/x2=(x1+x2)/(x1x2)=2
x1+x2=6/2=3
x1x2=3/2
1/x1+1/x2
=(x1+x2)/x1x2
=3/(3/2)
=2
Two
The sum of the two is equal to - B / A, and the multiplication of the two is C / A, so 1 / X1 + 1 / x2 = (x1 + x2) / X1 * x2
In other words, (* 2 - = 2) / C = 6
X & sup2; - 4Y & sup2; - 3x-2y + K can be divided into the product of two first-order factors
This paper is going to be 178; - 4Y & \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\; - 3x-2y + 2 = (X & #)
x²-4y²-3x-2y+2
=（x²-3x+9/4）-（4y²+2y+1/4）
=（x-3/2）²-（2y+1/2）²
=（X-3/2-2Y-1/2）（X-3/2+2Y+1/2）
=（X-2Y-2）（X+2Y-1）
K=2
If it can be divided into two linear factor products, it must be
[x²-3x+9/4]-[4y²+2y+1/4]=[x-3/2]²-[2y+1/2]²=[x-3/2+2y+1/2][x-3/x-2y-1/2]=[x+2y-1][x-2y-2]
By contrast, k = 2
Let the product of these two linear factors be: (x + 2Y + a) (x-2y + b)
After multiplication, the generation coefficient a + B = - 3, 2b-2a = - 2
The solution is: a = - 2, B = - 1, k = AB = 2
Solve the quadratic equation 4x-5x-7 = 0, y-4y-2 = 0 in Grade 9, Volume 1, using the matching method
The square of (m-2) - 4 = 0
The square of (2x-1) = 7 is made by direct square root
4X squared - 5x-7 = 0
x²-5x/4-7/4=0
x²-5x/4+25/64=137/64
(x-5/8)²=137/64
x=5/8+√137/8 x=5/8-√137/8
The square of y-4y-2 = 0
y²-4y+4=6
(y-2)²=6
y=2+√6 y=2-√6
The square of (m-2) - 4 = 0
(m-2)²=4
m-2=2 m-2=-2
m=4 m=0
Square of (2x-1) = 7
(2x-1)²=7
2x-1=√7 2x-1=-√7
x=1/2+√7/2 x=1/2-√7/2
It is known that f (x) is an odd function defined on the real number set R, and when x > 0, f (x) = x2-4x + 3, (I) find the value of F [f (- 1)]; & nbsp; & nbsp; (II) find the analytic expression of F (x); & nbsp; & nbsp; (III) find the minimum value of F (x) in the interval [T, t + 1] (T > 0)
(I) from the meaning of the problem, we can get: F (x) is an odd function defined on the real number set R, so f (- 1) = - f (1), and f (0) = 0. When x > 0, f (x) = x2-4x + 3, so f (1) = 0, so f (- 1) = 0. So f [f (- 1)] = f (0) = 0 4 ′ (II) if x ＜ 0, then - x ＞ 0, because
Let X1 x2 be the two roots of the equation 2x ^ 2-6x + 3 = 0, and find X1 ^ 3 + x2 ^ 3
The two roots of the equation 2x ^ 2-6x + 3 = 0 are the two roots of the equation 2x ^ 2-2x2-6x + 3 = 0, and we are the two roots of the equation 2x ^ 2-6x + 3 = the two roots of the equation 2x ^ 2-6x + 3 = the two roots of the equation 2x ^ 2-6x + 3, and we are the two roots of the equation, and we are the two roots of the equation 2x2x2x + x2 = -6 \\\\\\\ \\\\\\\2x11111; x2 = 6 \\\\\\\\\\\and
The equation (3x-2) & sup2; = 4 (x-3) & sup2;
Method 1: divide 9x & sup2; - 12x + 4 = 4x & sup2; - 24x + 365x & sup2; + 12x-32 = 0 (5x-8) (x + 4) = 0x = - 8 / 5 or x = - 4 on both sides. Method 2: square both sides (3x-2) & sup2; = 4 (x-3) & sup2; 3x-2 = ± 2 (x-3). When 3x-2 = 2 (x-3), the solution is x = - 4. When 3x-2 = - 2 (x-3), the solution is x = 8 / 5
Quadratic equation of one variable: the square of 3y-4y-1 = 0
3y^2-4y-1=0
a=3,b=-4,c=-1
△=b^2-4ac=16+12=28>0
The equation has two unequal real solutions
X1 = [- b-radical (b ^ 2-4ac)] / (2a) = (4-radical 28) / 6 = (2-radical 7) / 3
X2 = [- B + radical (b ^ 2-4ac)] / (2a) = (4 + radical 28) / 6 = (2 + radical 7) / 3
Formula method, it seems that there is no simple method
2 / 9 plus or minus 2 root sign 13
Can't formula method be used? Formula is king.
Given that the odd function y = f (x) defined on the real number set R satisfies f (x + 2) = - f (x), then the value of F (6) is
f(x+2)=-f(x)
f(x+4)=-f(x+2)=-(-f(x))=f(x);
That is to say, this function is a periodic function with a period of 4
And this function is odd
So f (0) = 0;
f(6)=f(2+4)=f(2+0)=-f(0)=0
0。
Let X1 and X2 be the two roots of the equation 2x & # 178; - 6x + 3 = 0
(2x-1)(x-3)=0
x1=1/2
x2=3