It is known that a, B and C are prime numbers greater than 3, and 2A + 5B = C (1). This paper proves that there is a positive integer n > 1, so that the sum of a + B + C of all three prime numbers a, B and C satisfying the problem is equal It is divisible by n; (2) Find the maximum value of N in the above question

Here we take two sets of values for analysis: (1) if a = 11, B = 5, then C = 22 + 25 = 47, a + B + C = 63 (2) if a = 13, B = 7, then C = 26 + 35 = 61, a + B + C = 81 ∵ the greatest common divisor (63,81) = 9 ∵ n. The maximum possible value is 9. We prove that: ∵ 2A + 5B = C ∵ a + B + C = a + B + 2A + 5B = 3A + 6B = 3 (a + 2b) | a + B + C let a

When a is 1,2,3,5 respectively, 2A + 1 is prime. What are these prime numbers?

3 5 7 11

A is a prime number, B is a positive integer. The square of (2a + b) is 509 (4a + 51B). Find the value of a and B

509 is a prime number, so if 509 (4a + 51B) is a square number, only 4A + 51B = 509 * a certain square number, then 2A + B = 509 * a certain number, then 4A + 51B can be divisible by 509, so 2A + B must be divisible by 509, then we can see that 51 (2a + b) = 102a + 51B must also be divisible by 509, then two formulas subtract to get 9

Let the two roots of the equation 2x & # 178; - 3x + 1 = 0 be X1 and X2, and find: 1. X1 & # 178; + x2 & # 178; 2. One part of X1 + one part of x2

Let the two roots of the equation 2x & # 178; - 3x + 1 = 0 be X1 X2, then X1 + x2 = - (- 3) / 2 = 3 / 2, X1 * x2 = 1 / 2
X1²+X2²=（X1+X2）²-2*X1 *X2=（3/2）²-2*（1/2）=5/4
. X1 / 1 + x2 / 1 = (x1 + x2) / (x1 * x2) = (3 / 2) / (1 / 2) = 3

Given that the function f (x) = the square of X + (M + 2) x-3 is even, then the zero point of F (x) is zero·····

Because it is an even function, there is no odd power of X in the expression of F (x), that is, M + 2 = 0, M = - 2
F (x) = x & sup2; - 3 = 0, x = ± √ 3 the zero point of function f (x) is ± √ 3

If the system of inequalities {2x-3 > = 0 x

Where is m

Change quadratic function y = - 1 / 2x ^ 2-2x + 3 into vertex form

1. The formula of y = the square of X - 6x + m gives y = (x-3) the square of - 9 + M
So - 9 + M = 1
m=10
2. The formula of y = the square of X + 2aX + 1 + 2a is y = (x + a) - the square of a + 1 + 2A
Get y = = (x + a) square + a square + 1
A0, the axis of symmetry is to the right of the Y axis
Square of a + 10
So the vertex is in the first quadrant
3. Because the axis of symmetry is a straight line, x = 2
So y = (X-2) ^ 2 + K
Substituting (0,3)
k=-1
So y = (X-2) ^ 2-1

How to decompose [x (1 + X '* x') - x '(1 + X * x)] / (1 + X * x) (1 + X' * x '), where * is a multiplication sign and X and X' are two unknowns

Write x * as y
The original formula = (x (1 + Y & # 178;) - Y (1 + X & # 178;) / (1 + X & # 178;) (1 + Y & # 178;)
=(x+xy²-y-x²y)/(1+x²)(1+y²)
=(x-y)(1-xy)/(1+x²)(1+y²)

Given the equation KX + M = (2k-1) x + 4 about X, when km takes what value: (1) the equation has unique solution (2) there are countless solutions (3) there is no solution

kx+m=(2k-1)x+4,
kx+m=2kx-x+4,
(k-1)x=m-4
(1)k-1≠0
K ≠ 1 equation has unique solution;
(2)k-1=0,m-4=0
Namely
K = 1, M = 4 have innumerable solutions;
(3)k-1=0,m-4≠0
K = 1, m ≠ 4, no solution

5 + 6 / (x-16) = (2x-1) / (x + 4) - (3x-1) / (4-x) to solve the equation

It should be the square of x minus 16. If it is, then: multiply both sides of the equation by X & # 178; - 16 to get 5 (X & # 178; - 16) + 6 = (2x-1) · (x-4) + (3x-1) · (x + 4) 5x & # 178; - 80 + 6 = (2x & # 178; - 9x + 4) + (3x & # 178; + 11x-4) - 74 = 2x solution to get x = - 37 test: when x = - 37, X & # 178; - 1