The following sets are represented by enumeration: ① a = {x | X & # 178; = 9} ② B = {x ∈ n | 1 ≤ x ≤ 2} ③ C = {x | X & # 178; - 3x + 2 = 0}

The following sets are represented by enumeration: ① a = {x | X & # 178; = 9} ② B = {x ∈ n | 1 ≤ x ≤ 2} ③ C = {x | X & # 178; - 3x + 2 = 0}

①A={x|x²=9}={-3,3}
②B={x∈N|1≤x≤2}={1,2}
③C={x|x²-3x+2=0}={1,2}
If you don't understand, please hi me, I wish you a happy study!

If the quadratic trinomial ax + 3x + 4 cannot decompose the factor in the range of real number, then the value range of a is
A. 0 ＜ a ＜ 16 / 9, and a is less than 0 (this is impossible) B, a ≠ 0 (this seems not right) C, a ＞ 16 / 9 D, a is less than 3 / 4, and a ≠ 0 is mainly uncertain whether it is C or D Better have a process Of course, it doesn't matter Be sure to be right

Aunt is C

A = {x | X & # 178; = 9} B = {x ∈ n | 1 ≤ x ≤ 2} C = {x | X & # 178; - 3x + 2 = 0}

A={-3,3}
B={1,2}
C={1,2}

If the quadratic trinomial ax & # 178; + 3x + 4 = O cannot decompose the factor in the range of real number, then the value range of a is?

Because the quadratic trinomial ax & # 178; + 3x + 4 = O cannot decompose the factor in the range of real number
So b2-4ac

Finding the minimum value of quadratic function y = ax & # 178; + 4A & # 178; X + 1 (- 2 ≤ x ≤ 4)

Axis of symmetry x = - 4A
The discussion is divided into four situations
1.a4
When the opening of parabola is downward and x = - 2, Ymin = 4a-8a & # 178; + 1
2.-1≤a

Ax & sup2; + 3x + 4 can not decompose the factor in the range of real number, then the value range of a is
A. 0 ＜ a ＜ 9 out of 16, and a ＜ 0 B.A ≠ 0 C.A ＞ 9 out of 16, D.A ＜ 4 out of 3, and a ≠ 0 will be needed today
The quadratic equation of one variable is
A. 4X & sup2; = 3Y & sup2; b.x (x + 1) = 5x & sup2; - 1 C. radical x-3 = 5x & sup2; - radical 6 D.X & sup2; 1 / 2 + 3x-1 = 0

1.
We can't decompose the factor, that is, ax & sup2; + 3x + 4 = 0 has no solution
So 3 & sup2; - 4A * 4 < 0
That is, a > 9 / 16
Choose C
2. Univariate quadratic equation means that there is only one unknowns, and the highest degree of unknowns is 2
A: It's a quadratic equation of two variables
B: Right
C: The highest order must be greater than 2
D: Reducibility

If the quadratic function f (x) = x & # 178; + ax + B passes through the point (1,0) and its minimum value is the same as the function g (x) = - X & # 178; - 2
The maximum value of X is the same, 1. Find the expression of function f (x), 2. Find the maximum and minimum value of function f (x) - G (x) on [- 2,2]!

(1) When the function g (x) = - x ^ 2-2 takes the maximum value, x = 0, so f (x) = x ^ 2 + ax + B takes the minimum value at x = 0, then a = 0, and the parabola passes through the point (1,0), substituting into b = - 1, so the analytic expression of F (x) is f (x) = x ^ 2-1. (2) f (x) - G (x) = (x ^ 2-1) - (- x ^ 2-2) = 2x ^ 2 + 1, its axis of symmetry

In order to factorize the quadratic trinomial x2-5x + P in the range of integers, the value of integer P can be () A. 2 b. 4 C. 6 D. innumerable

If the quadratic trinomial x2-5x + P can be decomposed, there must be: 25-4p ≥ 0, that is, P ≤ 254, which can be factorized within the range of integers. Therefore, as long as P can be decomposed into two integers, and the sum is - 5, such numbers have arrays, so there can be innumerable values of integer P. therefore, choose D

It is known that the quadratic trinomial x ^ 2 + ax-18 can decompose factors in the range of rational numbers, find the possible values of integer a, and decompose all factors

∵18=1×18=2×9=3×6
∴
x^2+ax-18=(x-18)(x+1)=(x+18)(x-1)=(x-2)(x+9)=(x+2)(x-9)=(x-3)(x+6)=(x+3)(x-6)
The possible values of integer a are: - 17,17,7, - 7,3, - 3

It is known that the quadratic trinomial x ^ 2 + ax-18 can decompose the factor in the range of rational number, find the possible value of integer a, and decompose the factor

Let (x + m) (x + n) = x & # 178; + (M + n) x + Mn = x & # 178; + ax-18m + n = amn = - 18 = - 1 * 18 = - 2 * 9 = - 3 * 6 = - 6 * 3 = - 9 * 2 = - 18 * 1, so there are six a = ± 17, ± 7, ± 3, you can write them yourself, such as X & # 178; - 7x-18 = (X-9) (x + 2) x & # 178; + 3x-18 = (x + 6) (x-3)