Let P: x ^ 2-8x-20 > 0, Q: x > 1 + A or X be known

Let P: x ^ 2-8x-20 > 0, Q: x > 1 + A or X be known

The negation of Q:
X is less than or equal to (1 + a) and X is greater than or equal to (1 + a),
We get: x = 1 + a
The negation of P: x ^ 2-8x-20 is less than or equal to 0
(x + 2) (X-10) less than or equal to 0
-2 less than or equal to x less than or equal to 10
Because the negation of Q is a sufficient and unnecessary condition for the negation of P
So: - 2 less than or equal to (1 + a) less than or equal to 10
A belongs to [- 3,9]

The imaginary number X-2 + Yi is known, where x and y are real numbers. When the module of the imaginary number is 1, the value range of X / y is obtained

Shilling X / y = T. then x = YT, bring in the formula to find the module. You will find a quadratic equation. Then use the discriminant to have roots in the range of real numbers to find the range of T. remember to discuss the case of T = 0 before. Then you can find the answer

Given that the module of imaginary number (X-2) + Yi (x, y ∈ R) is 1, then the value range of x ^ 2 + y ^ 2-2 is?

Set
x = 2 + sint
y = cost
x^2+y^2-2 = 5 + 4sint - 2 = 3+4sint
Because - 1

Using the relationship between root and coefficient, if the two equations are 2 and 4, then the equation is?
I'll tell you in detail from every aspect

x²-6x+8-0