Given the complete set u = {1,2,3,4,5} a = {x | x ^ 2-5x + q = 0}, find the complement of a (how to calculate the value of X in set a,

Given the complete set u = {1,2,3,4,5} a = {x | x ^ 2-5x + q = 0}, find the complement of a (how to calculate the value of X in set a,

Three cases
(1)A=Φ
(2)A={2,3}
(3)A={1,4}
Note that there is a complete set limitation, so,
The real root of the equation (if any) can only be in 1,2,3,4,5

Let u = R, a = {x | x ^ - 5x + 6 > 0}, B = {x | x + 3 | ≥ 6}, find the complement a,
Answer correctly and give two absolute values

If set a: x-5x + 6 = (X-2) (x-3) > 0, then a = {x | x ＞ 3 or X ＜ 2}, then the complement of a = {x | 2 ≤ x ≤ 3} (2) if set B - 6 ≤ x + 3 ≤ 6, then - 9 ≤ x ≤ 3, then the intersection of complement a B = {x | 2 ≤ x ≤ 3}

We know that the complete set {1,2,3,4,5}, the set a = {X / x ^ 2-5x + M =}, B = {X / x ^ 2 + NX + 12 = 0}, and (the complement of a) u b = {1,3,4,5}, and find the complete set of M + n
We know the complete set {1,2,3,4,5}, the set a = {X / x ^ 2-5x + M =}, B = {X / x ^ 2 + NX + 12 = 0}, and (the complement of a) u b = {1,3,4,5}, find the value of M + n

(A's complement) ∪ B = {1,3,4,5}, explain 2 &; a's complement, then 2 ∈ a substitute x = 2 into a, get: 4-10 + M = 0, then M = 6, solve the equation x ^ 2-5x + 6 = 0, (X-2) (x-3) = 0, x = 2, or x = 3, then a = {2,3}, then a's complement = {1,4,5}, and (A's complement) ∪ B = {1,3,4,5}, explain 3 ∈ B substitute x = 3 into B

x. If y is an integer and 9x + 5Y, 10x + KY can be divided by 11, then k =?

9x + 5Y can be divided by 11 (9x + 5Y) * 10 = 90x + 50y can be divided by 11
10x + KY can be divisible by 11 (10x + KY) * 9 = 90x + 9ky can be divisible by 11
XY is an integer, so 50-9k is divisible by 11
So k = 8 19 30
K = 11n + 8, n is a natural number

If x, y are integers, and one of 2x + 3Y, 9x + 5Y can be divisible by 17, then the other can also be divisible by 17

It is proved that: let u = 2x + 3Y, v = 9x + 5Y. If 17 | u, eliminate y from the above two formulas, we get 3v-5u = 17x. ① so 17 | 3V. Because (17, 3) = 1, so 17 | V, that is 17 | 9x + 5Y. If 17 | V, we also know 17 | 5u from formula ①. Because (17, 5) = 1, so 17 | u, that is 17 | 2x + 3Y

If one of the integers 2x + 3Y, 9x + 5Y can be divisible by 17, then for the same X, y, the other can be divisible by 17. (x, y are integers)
Speed, thanks

Let u = 2x + 3Y, v = 9x + 5Y
(1) If 17 | u, y is deleted from the above two formulas,
The results show that 3v-5u = 17x
So 17:3v
Because (17,3) = 1,
So 17|v,
That is 17|9x + 5Y
(2) If 17|v, we can also know 17|5u from Formula 1
Because (17,5) = 1,
So 17 u,
That is 17|2x + 3Y

Prove that: (x ^ 2-xy + y ^ 2) ^ 3 + (x ^ 2 + XY + y ^ 2) ^ 3 can be divisible by 2x ^ 2 + 2Y ^ 2

There is a formula
a³+b³=(a+b)(a²-ab+b²)
so
(x^2-xy+y^2)^3+(x^2+xy+y^2)^3
=(x^2-xy+y^2+x^2+xy+y^2)((x^2+xy+y^2)^2-(x^2-xy+y^2)*(x^2+xy+y^2)+(x^2+xy-y^2)^2)
=(2x^2+2y^2)((x^2+xy+y^2)^2-(x^2-xy+y^2)*(x^2+xy+y^2)+(x^2+xy-y^2)^2)
The factor 2x ^ 2 + 2Y ^ 2
therefore
(x ^ 2-xy + y ^ 2) ^ 3 + (x ^ 2 + XY + y ^ 2) ^ 3 can be divisible by 2x ^ 2 + 2Y ^ 2

A number divided by 2, the remainder is 1, divided by 3, the remainder is 2, divided by 4, the remainder is 3, divided by 5, the remainder is 4, divided by 6, the remainder is 5. This number is divided by 7

2. The least common multiple of 3, 4, 5 and 6 is 60
60-1=59
Adding 60 each time can satisfy the least common multiple of 2, 3, 4, 5 and 6, and then consider the multiple of 7
59+60=119=7*17
So the answer is 119

Finding the negative integer solution of 3 / 2 (x-3) ≤ 6 / (5x-4) - 1
Wrong number on 2 (x-3) / 3 ≤ (5x-4) - 1 / 6

In the same way, we need to make it a little simpler. First, we get the following results:
2x-6=1/2
Therefore, there is no x value of negative integer in line with the meaning of the problem,
Another way is the answer
Oh, according to the fourth floor, the right side is [(5x-4)] - 1 / 6

x+24=4.5X

x+24=4.5x
24=4.5x-x
24=3.5x
x=48/7