Let a be a prime number, B be a positive integer, and 9 (2a + b) (2) = 509 (4a + 511b) How to solve the problem 509 is a prime number, 2A + B is an integral multiple of 509, let 2A + B = 509k (k is an integer) Substituting, 9kk * 509 * 509 = 509 (509B + 2 * 509k) 9KK=b+2K=509K-2a+2K 2a=(511-9K)k Because a is a prime number, k = 1 or 511-9k = 1 (k is not equal to integer, exclusive) or K = 2 (a = 493 = 17 * 29, non prime, exclusive) or 511-9k = 2 (k is not integer, exclusive) So k = 1, a = 251, B = 509-2 * 251 = 7 There are a few things I don't understand 2a=(511-9K)k Why is k = 2 or (511-9k) = 2 or K = 1 or (511-9k) = 1 because a is prime How to verify "when 3 ≤ x ≤ 56, a must be a composite number", just write the above formula Is it unnecessary to say (511-9x) = 1 or = 2?

Let a be a prime number, B be a positive integer, and 9 (2a + b) (2) = 509 (4a + 511b) How to solve the problem 509 is a prime number, 2A + B is an integral multiple of 509, let 2A + B = 509k (k is an integer) Substituting, 9kk * 509 * 509 = 509 (509B + 2 * 509k) 9KK=b+2K=509K-2a+2K 2a=(511-9K)k Because a is a prime number, k = 1 or 511-9k = 1 (k is not equal to integer, exclusive) or K = 2 (a = 493 = 17 * 29, non prime, exclusive) or 511-9k = 2 (k is not integer, exclusive) So k = 1, a = 251, B = 509-2 * 251 = 7 There are a few things I don't understand 2a=(511-9K)k Why is k = 2 or (511-9k) = 2 or K = 1 or (511-9k) = 1 because a is prime How to verify "when 3 ≤ x ≤ 56, a must be a composite number", just write the above formula Is it unnecessary to say (511-9x) = 1 or = 2?

Let 2A + B = 509x (x ≥ 1), then 9 × (509x) ^ 2 = 509 × (4a + 2B + 509B) 9 × 509 × x ^ 2 = 2 × 509x + 509B 9 × x ^ 2 = 2x + B = x (9x-2) substitute B into 2A + B = 509x, then 2A + X (9x-2) = 509x, that is, a = x (5

Let a be prime numbers, B and C be positive integers, and the square of 9 (2a + 2b-c) = 509 (4a + 1022b-511c) and B-C = 2, then the value of a (B + C) can be obtained

Let a be prime, B be positive integer, and 9 (2a + b) & sup2; = 509 (AA + 511b) find the value of a and B

When B is even, the equation is even. When B is odd, the equation is odd. When a is even, a is prime. A = 2 (3 * 4 + 3 * b) ^ 2 = 509 (4 + 511b) 9b ^ 2 + 72B + 144 = 2036 + 260099b 9b ^ 2-260027b-1892 = 0 B (9b-260027) = 1892 B is a positive integer, so there is no solution

5-3/8” x 1-1/2” x 5-3/4”.
It's not a multiply sign in the middle, it should be*

5-3/8 x【（ 1-1/2） x 5-3/4】
=5-3/8 x【1/2 x 5-3/4】
=5-3/8 x【2.5-3/4】
=5-3/8×7/4
=5-21/32
=4 and 11 / 32

Solving the equation of one degree with one variable
,10(x+30)=15x
,2.8(x+24)=3(x-24)
If the brackets are removed, the transferred items will be merged, and the coefficients of the similar items will be converted into one
What's more, I would like to ask how to turn coefficient into one?
PS: let's turn coefficients into one. It's better to list a few questions to understand

1.10 (x + 30) = 15x get: 10x + 300 = 15x get: 10x -- 15x = -- 300 get: - 5x = -- 300 get: x = 602.2.8 (x + 24) = 3 (X -- 24) get: 2.8x + 67.2 = 3x -- 72 get: 2.8x -- 3x = -- 72 -- 67.2 get: - 0.2x = -- 139.2

Implicit function derivation exercises
It is known that f (x) is a second-order differentiable single valued function, f (1) = 0, f '(1) = 5, F "(1) = 7. Y = y (x) satisfies the equation: F (x + y) = XY + X. find: dy / DX (x = 0), d2y / DX2 (x = 0),

Let x = 0 in F (x + y) = XY + X, then f (y) = 0. Because f (x) is a second-order differentiable single valued function, f (1) = 0, so y = 1. Then y = 1 when x = 0. If f (x + y) = XY + X is left and right, then f '(x + y) * (1 + y') = y + XY '+ 1

Factorization x ^ 3 + 8y ^ 3 + 125Z ^ 3-30xyz

Given the absolute value of (2x-1) + the absolute value of (3y-2) = 0, find X and y

The absolute value of (2x-1) + the absolute value of (3y-2) = 0
The absolute value of (2x-1) is 0, and the absolute value of (3y-2) is 0
x=1/2,y=2/3

If y = f (x) is a quadratic function and f (x) + F (2x) = 5x ^ 2 + 3x + 2, f (x)=_____ .
How do you do it?

Let f (x) = ax ^ 2 + BX + C. A ≠ 0
f(x) + f(2x)
= 5ax^2 + 3bx + 2c
= 5x^2 + 3x + 2
a = b = c = 1
f(x) = x^2 + x + 1

The known function y = (2m + 1) x + m + 3, if this function is a linear function and Y decreases with the increase of X, the value range of M is obtained

Function y = (2m + 1) x + m + 3 if the function is linear
Then 2m + 1 ≠ 0, M + 3 ≠ 0
Y decreases with the increase of X
2m+1