If K is an integer, we know that the univariate quadratic equation kx2 + (2k-1) x + k-1 = 0 has only integer roots, then K=______ ．

If K is an integer, we know that the univariate quadratic equation kx2 + (2k-1) x + k-1 = 0 has only integer roots, then K=______ ．

Kx2 + (2k-1) x + k-1 = 0, ∧ (KX + k-1) (x + 1) = 0, ∧ X1 = - 1, X2 = 1 − KK = 1k-1, because there is only an integer root, the K with 1 − KK as an integer can be taken as ± 1

It is known that set a = {x | x = 2K, K belongs to integer}, set B = {x | x = 2k-1, K belongs to integer}
Finding union of sets a and B
A and B = {x | x = 2K or 2k-1, K belongs to integer} 1
A and B = {x | x = k, K belongs to integer} 2
A and B = integer 3
How do you get the second step from the first step?

X = 2K, K is an integer, the number divisible by 2 is even, then x is even
If x = 2k-1, K is an integer, then x is odd
Odd and even numbers together are integers

Let a = {x | x = 2k-1, K be an integer}, B = {x | x = 2K + 1, K be an integer}, then the relationship between a and B is?
What's the solution
If a = {x | x = n / 2-1 / 3, n is an integer}, B = {x | x = P / 2 + 1 / 6, P is an integer}, then what is the relationship between a and B? B = a, too?

These two sets are equal sets, both of which represent odd numbers
You see: 2k-1 = 2 (k-1) + 1
Since K is all integers, there is no difference between k-1 and K
Right. P / 2 + 1 / 6 = (P + 1-1) / 2 + 1 / 6 = (P + 1) / 2-1 / 3
Since N and P are integers, then p + 1 and N are indistinguishable