If we know that x is an integer and 2 / (x + 3) + 2 / (3-x) + (2x-18) / (the square of X-9) is an integer, then the sum of all the values of X that meet the condition is? See, it's (2x-18) / (x squared-9), not (2x + 18) / (x squared-9),

If we know that x is an integer and 2 / (x + 3) + 2 / (3-x) + (2x-18) / (the square of X-9) is an integer, then the sum of all the values of X that meet the condition is? See, it's (2x-18) / (x squared-9), not (2x + 18) / (x squared-9),

2 / (x + 3) + 2 / (3-x) + (2x-18) / (the square of X-9) = k, K is an integer
It is reduced to k * x ^ 2-2x + 30-9k = 0
When k = 0, it is a univariate equation, x = 15
When k is not equal to 0:
Only when △ = B ^ 2-4ac ≥ 0, that is, 9K ^ 2-30k + 1 ≥ 0, can the equation be solved
The sum of the two is: - B / a = 2 / K
9K ^ 2-30k + 1 ≥ 0, we can get the integer between K value range (- ∞, 0] ∪ [4, + ∞)
When k is the opposite, 2 / K will cancel each other
Therefore, the sum of possible values of X is 15 + 2 / - 1 + 2 / - 2 + 2 / - 3 = 34 / 3

Given that x is an integer and 2 / (x + 3) + 2 / (3 + x) + (2x-18) / (x.x-9) is an integer, find all the conditions
emergency
It is known that x is an integer and [2 / x + 3] + [2 / 3-x] + [2x + 18 / X & # 178; - 9]
Is an integer,
Find the sum of all qualified x values.

[2/(x+3)]+[2/(3-x)]+[(2x+18)/(x²-9)]
=2/(x-3)
∵ x is an integer and 2 / (x-3) is also an integer
Ψ x-3 = ± 2 or ± 1
Then x = 5 or 1 or 4 or 2
5+1+4+2=12

Let u = R, a = {x | x2 + PX + 12 = 0}, B = {x | x2-5x + q = 0}, if (∁ UA) ∩ B = {2}, a ∩ (∁ UB) = {4}, find a ∪ B

∵4∈A2∈B⇒p＝−7q＝6∴A={3，4}，B={2，3}∴A∪B={2，3，4}

Let u = {x | x ≤ 5, and X ∈ n}, a = {x | X's Square - 5x + Q}, B = {x | X's square + PX + 12 = 0}, if Cu (a ∩ b) = {1,2,3,5} find a ∪ B

According to u = {x | x ≤ 5, and X ∈ n}, u = {1,2,3,4,5} is obtained
Cu (a ∩ b) = {1,2,3,5} we get that both a and B contain {4}
If one root of the square of X - 5x + q = 0 is 4, then the other root is 1 (the sum of the two is 5), that is, a = {1,4}
If one root of X | x square + PX + 12 = 0 is 4, then the other root is 3 (the product of the two is 12), that is, B = {3,4}
Then a ∪ B = {1,3,4}