Let f (x) = x2-3x + 3, then f (a) + F (- a) equals?

Let f (x) = x2-3x + 3, then f (a) + F (- a) equals?

The function f (x) = x2-3x + 3,
Then f (a) + F (- a) = a ^ 2-3a + 3 + (- a) ^ 2 + 3A + 3 = 2A ^ 2 + 6

Let f (- x) = x2 + 3x + 1, then f (x + 1)=

F (- x) = x2 + 3x + 1, change - x to X
So f (x) = (- x) ^ 2 + 3 (- x) + 1 = f (- x) = x ^ 2-3x + 1
So f (x + 1) = (x + 1) ^ 2-3 (x + 1) + 1 = x ^ 2-x-1

If the definition field of function f (x) = root sign (2 ^ (x ^ 2-2ax-1)) is r, find the value range of real number a

The exponential function is greater than 0
SO 2 ^ (x ^ 2-2ax-1) > 0
So √ [2 ^ (x ^ 2-2ax-1)] must be meaningful
So a takes any real number

Let a = {x | X & # 178; + 3x-10

x=3
Multiply a set by cross to get (X-2) (x + 5)

Factorization of higher order terms
Our teacher said, for example, a formula with five terms has five roots, which can be multiplied by five formulas. What's more advanced than cross multiplication

Factor theorem is one of the corollaries of remainder theorem: if the polynomial f (a) = 0, then the polynomial f (x) must contain the factor x-a. conversely, if f (x) contains the factor x-a, then f (a) = 0. It is easier to decompose the factor by combining the factor theorem with the coefficient method

（1） （5m²＋3n²）²－（3m²＋5n²）²
(2) XY & # 179; - 2x & # 178; Y & # 178; + X & # 179; Y (this is 3)
（3） （x²＋4y²）²－16x²y²

（5m²＋3n²）²－（3m²＋5n²）²=【（5m²+3n²）-（3m²+5n²）】【（5m²+3n²）+（3m²+5n²）】=（2m²-2n²）（8m²+8n²）=16...

Math problem! Factorization! Urgency! Process!
①x²-x-6
②x²+x-12
③-x²-5x-4
④x²+3xy-28y²
⑤（a+b）²-4（a+b）-12
⑥（x²-x）²-4

①x²-x-6=(x-3)(x+2)
②x²+x-12=(x+4)(x-3)
③-x²-5x-4=-(x+4)(x+1)
④x²+3xy-28y²=(x+7y)(x-4y)
⑤（a+b）²-4（a+b）-12=(a+b-6)(a+b+2)
⑥（x²-x）²-4
=(x²-x-2)(x²-x+2)
=(x-2)(x+1)(x+2)(x-1)

On the factorization of mathematical problems, we need to process
6a³-9a²b²c
a²-ab²
-x²+xy
4a²-8ab+4a
21xy-14xz+35x²
2a（x-y）-6b（y-x）
12a(x²+y²)-18b（x²+y²）
Please

The following topics are in the order: 1, 6a-6a & \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(5) 21xy-14xz + 35x # 178

Five mathematical problems about factorization
1. If a, B, C and D are four positive numbers whose product is 1, then the minimum value of the algebraic formula A & sup2; + B & sup2; + C & sup2; + D & sup2; + AB + AC + AD + BC + BD + CD is?
2. Positive integer n makes 2n + 1 and 3N + 1 square numbers. Is 5N + 3 prime?
3. If integers a and B satisfy 6ab-9a + 10B = 303, find a + B
4. Factorization: the fifth power of a + A + 1
5. Given that a and B are rational numbers and satisfy 2A + A & sup2; + A & sup2; B & sup2; + 2 + 2Ab = 0, then the value of a + B is zero

1.
ABCD = 1, a & sup2; + B & sup2; + C & sup2; + D & sup2; + AB + AC + AD + BC + BD + CD ≥ 10 [(ABCD) ^ 5] to the power of 10 = 10
Therefore, the minimum value of a & sup2; + B & sup2; + C & sup2; + D & sup2; + AB + AC + AD + BC + BD + CD is 10, which is obtained when a = b = C = D = 1
two
Positive integer n makes 2n + 1 and 3N + 1 square, then n can only be 1, 5N + 3 = 8, not prime
three
6ab-9a+10b-15=(3a+5)(2b-3)=303-15=288=2^5×3²
If 3A + 5 is not a multiple of 3 and 2b-3 is odd, then 2b-3 can only be 9, 3A + 5 = 2 ^ 5 = 32
Then we can get: a = 9, B = 6, a + B = 15
four
a^5+a+1
=a^5-a²+a²+a+1
=a²(a³-1)+(a²+a+1)
=a²(a-1)(a²+a+1)+(a²+a+1)
=(a²+a+1)(a³-a²+1)
five
2a+a²+a²b²+2+2ab=0
(a+1)²+(ab+1)²=0
So a + 1 = 0, AB + 1 = 0,
The solution is: a = - 1, B = 1
So a + B = 0

Factorization: (1) 2x (a-b) - (B-A); (2) x (x + 4) + 3; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (3) x2-2x + 1-y2

(1) Original formula = 2 & nbsp; X & nbsp; (a-b) + (a-b) = (a-b) (2 & nbsp; X + 1); (2) original formula = x2 + 4x + 3 = (x + 1) (x + 3); (3) original formula = (x-1) 2-y2 = (x + Y-1) (x-y-1)