If a

If a

X is greater than - 2 / (3-A)
Let him be a hot substitute
（a-3）x＜2,x＞2/（a-3）
It is known that the solution set of inequality ax-5x2-a < 0 about X is m. if 3 ∈ m and 5 ∉ m, then the value range of real number a is______ .
If 3 ∈ m and 5 ∉ m, the solution set of inequality ax-5x2-a ＜ 0 about X is m, so there are & nbsp; 3 · a-59-a ＜ 0 & nbsp; & nbsp; & nbsp; & nbsp; ① 5 · a-525-a ≥ & nbsp; 0 or & nbsp; 25-A & nbsp; = 0 & nbsp; & nbsp; & nbsp; ②, a ＜ 53 & nbsp;, & nbsp; or a ＞ 91 ≤ a ＜ 25
Given that the solution of inequality x ^ 2 + ax + 1 ≥ 0 is all real numbers, find the value range of A
△=a^2-4
The minimum value of x ^ 2 + ax + 1 is 4-A ^ 2
4-a^2>=0
A ^ 2 = a > = - 2
If the inequality ax ^ 2-ax + 2
1.a=0
Obviously wrong
2.a＜0
Δ=a²-4×a×2≤0
a(a-8)≤0
0≤a≤8
therefore
There is no solution to the problem
When a & gt; 0, the opening of quadratic function is upward, so there is no such a. when a = 0, the original inequality becomes 2 & lt; = 0, which is obviously not tenable. When a & lt; 0, @ = A & # 178; - 4 × a × 2 is greater than 0, Δ is tenable
Finding the range of function y = √ 1-2 Λ x
Because 2 ^ x > 0
So 1-2 ^ x = 0
So 0
Quadratic function f (x) = AX2 + BX + C "for any real number x, f (x) is greater than or equal to 0"
Is a > 0, bsquare-4ac = 0 or bsquare-4ac less than or equal to 0? Why?
The quadratic function corresponding to the coordinate axis is a quadratic curve, f (x) > = 0, indicating that the curve is above or tangent to the X axis, and the opening is upward, that is, a > 0. In addition, as long as there is only one or no intersection point between the curve and the X axis, that is, there is only one real root or no real root, the necessary and sufficient condition is B ^ 2-4ac
a> 0 means that the image opening of this function is upward. As long as there is no or only one intersection point between its image and X axis, we can have f (x) greater than or equal to 0 for any real number X. in this way, b-square-4ac
The range of function y = (1 / 2) ^ x ^ 2
Because x ^ 2 ≥ 0
So y = (1 / 2) ^ x ^ 2 ≤ (1 / 2) ^ 0 = 1
That is, the value range is (0,1]
This problem is easy to make mistakes. Be careful!
As shown in the figure, the image of the quadratic function y = AX2 + BX + C (a is not equal to 0) passes through the point (- 1,2) and is consistent with X
As shown in the figure, the image of the quadratic function y = AX2 + BX + C (a ≠ 0) passes through the point (- 1,2), and the abscissa of the intersection point with the X axis is X1 and X2 respectively, where - 2
Three messages can be seen in the picture
(1)a-b+c=2
(2)4a-2b+c
The range of function y = 1 / 2 ^ X-1
If the title is:
y=1/(2^x-1)
The inverse method is used
1=y2^x-y
y2^x=1+y
2^x=(y+1)/y>0
(y+1)y>0
y> 0 or Y
2^x>0
1/2^x>0
1/2^x-1>-1
y> - 1 this is the range
The range is (- 1, + infinity)
The image of quadratic function y = AX2 + BX + C (a ≠ 0) is shown in the figure. If M = a + B-C, n = 4a-2b + C, P = 2a-b, then in M, N, P, the number less than 0 has ()
A. 3 B. 2 C. 1 D. 0
When x = - 2, y = 4a-2b + C ＜ 0, n = 4a-2b + C ＜ 0, ∵ - B2A ＞ - 1, B2A ＜ 1, ∵ a ＜ 0, ∵ B ＞ 2a, ∵ 2a-b ＜ 0, P = 2a-b ＜ 0, then m, N, P are less than 0