Solving inequality 2x & # 178; + kx-k ≤ 0 on X I see the classification of solutions i) Δ = k ^ 2 + 8K < 0, i.e. - 8 < K < 0 In this case, 2x2 + kx-k = 0 has no root, f (x) = 2x2 + kx-k has no intersection with X axis, so the solution of 2x2 + kx-k ≤ 0 is an empty set II) △ = k ^ 2 + 8K = 0, i.e. k = - 8 or K = 0 At this time, f (x) = 2x2 + kx-k and X axis have an intersection, that is - K / 4, so the solution of 2x2 + kx-k ≤ 0 is {2,0} III) △ = k ^ 2 + 8K > 0, that is K0 In this case, f (x) = 2x2 + kx-k and X-axis have two intersections, which are x1, X2 respectively, So the solution of 2x2 + kx-k ≤ 0 is {x | x1

Solving inequality 2x & # 178; + kx-k ≤ 0 on X I see the classification of solutions i) Δ = k ^ 2 + 8K < 0, i.e. - 8 < K < 0 In this case, 2x2 + kx-k = 0 has no root, f (x) = 2x2 + kx-k has no intersection with X axis, so the solution of 2x2 + kx-k ≤ 0 is an empty set II) △ = k ^ 2 + 8K = 0, i.e. k = - 8 or K = 0 At this time, f (x) = 2x2 + kx-k and X axis have an intersection, that is - K / 4, so the solution of 2x2 + kx-k ≤ 0 is {2,0} III) △ = k ^ 2 + 8K > 0, that is K0 In this case, f (x) = 2x2 + kx-k and X-axis have two intersections, which are x1, X2 respectively, So the solution of 2x2 + kx-k ≤ 0 is {x | x1

-B / 2a, do you understand? It's the abscissa of the pocket of quadratic function
-K / 4 is the abscissa of X, at the axis of symmetry
Axis of symmetry, - B / 2A
High school mathematics, if passing through point (1,2) can make two tangents of the square of circle x + y + 4kx + 2Y + K, then the value range of K is ()
The title should be x ^ 2 + y ^ 2 + 4kx + 2Y + k = 0
If there are two tangents, then the point (1,2) is outside the circle, that is, 1 + 4 + 4K + 4 + K ^ 2 > 0
That is, K ^ 2 + 4K + 9 ＞ 0, and △ = 16-36 ＜ 0, K ∈ R
Solving inequality x ^ 2-2x + 1-A ^ 2 > = 0 solving inequality 2x ^ 2 + ax + 2 > 0
1. (x-1-a) (x-1 + a) ≥ 0 can be obtained from X & # 178; - 2x + 1-A & # 178; ≥ 0, and the point equal to 0 is x = 1 + A or x = 1-A
When a > 0, 1-A < 1 + A, then the solution of the original inequality is X1 + a
When A0, the discriminant △ = A & # 178; - 16 = (a + 4) (A-4)
Because f (x) = 2x & # 178; + ax + 2, the image is a parabola with the opening upward
When the solution set of - 40 is x [- 4 + √ (A & # 178; - 16)] / 4
When a = - ± 4, △ = 0, the solution set of 2x & # 178; + ax + 2 > 0 is x ≠ ± 1
From X & # 178; - 2x + 1-A & # 178; = 0, (x-1-a) (x-1 + a) = 0
∴x1=1+a x2=1-a
If a > 0, the solution set of the inequality is: {x | x ≥ 1 + A, or X ≤ 1-A};
If A0
△=a²-8
If a > 2 √... Expand
From X & # 178; - 2x + 1-A & # 178; = 0, (x-1-a) (x-1 + a) = 0
∴x1=1+a x2=1-a
If a > 0, the solution set of the inequality is: {x | x ≥ 1 + A, or X ≤ 1-A};
If A0
△=a²-8
If ｛ or ｛ - 2, ｛ - A, ｛ - 2, ｛ - A, ｛ - A, ｛ - A, ｛ - A, ｛ - A, ｛ - A, ｛ -
P (x, y) is the moving point of the circle x ^ 2 + y ^ 2 = 2Y. Find the value range of Y / x + 2
x^2+(y-1)^2=1
So let x = Sina, y = 1 + cosa
Let t = Y / (x + 2)
=(cosa+1)/(sina+2)
tsina+2t=cosa+1
tsina-cosa=1-2t
√ (T ^ 2 + 1) sin (a - φ) = 1-2T
-1
Solving the square of inequality 2x + ax + 2 > 0
2. When we ask the value of M, the square of the equation x - 2mx + 2m + 3 = 0 has two negative roots
It's too much trouble to remind you. I'm too lazy to solve it
Question 1: first, the opening of function curve is upward. There is a possibility that deta is greater than 0. In this case, X is all real numbers. In the second case, X is equal to 0. In this case, except the root, all solutions of X are solutions. In the third case, X is less than 0. In this case, X is less than the small root and greater than the big root
The second problem: is the use of Weida's theorem, the sum of two is less than 0, the product of two is greater than 0, that is, 2 / 2m is less than 0, 3 / (- 2m) is greater than 0, the value range of M is obtained
Given the point P (1,2) and the circle C: x ^ 2 + y ^ 2 + KX + 2Y + K ^ 2 = 0, there are two tangents through P as C, then the value range of K is?)
(x-k/2)^2+(y+1)^2=1-3k^2/4
There are two tangent lines
So point P must be outside the circle,
Substituting point (1,2) into greater than 0
1^2+(2)^2+k+2*2+k^2>0
k^2+k+9>0
equation holds good under all circumstances
So long as the equation is a circle
Then 1-3k ^ 2 / 4 = R ^ 2 > 0
3k^2/4
(X+K/2)^2+(Y+1)^2=1-3K^2/4
Center: the distance from C (- K / 2, - 1) to p is greater than R
（1+K/2）^2+(2+1)^2>1-3K^2/4
k^2+k+9>0
(K+1/2)^2+35/4>0
R^2=1-3K^2/4>0
-2√3/3K0
k^2+k+9>0
Heng was founded,
So K can be any number
X ^ 2 + y ^ 2 + KX + 2Y + K ^ 2 = 0 (x + K / 2) ^ 2 + (y + 1) ^ 2 = 1-3k ^ 2 / 4 after simplification
Let p be less than the radius from the center of the circle
Solving inequality (1) ax ^ 2-A ^ 2x
1. When a = 0, the solution set of inequality is empty
When a is not equal to 0, consider the image of y = ax ^ 2-A ^ 2x = ax (x-a)
When A0, and the intersection point with X axis is 0 and a, the solution is x0
When a > 0, X (x-a)
1. When a = 0, the solution set of the inequality is r.
When a is not equal to 0, consider the image of y = ax ^ 2-A ^ 2x = ax (x-2a).
When A0, the beginning of the image is upward, and the intersection with the X axis is 0 and 2a, so the solution is 0
The solution of inequality (A-X) (x ^ 2-x-2) about x > 0, where constant a is a real number
Because two polynomials must have the same sign:
1：(a-x)>0;(x^2-x-2)>0；
The solution is X
(a-x)(x^2-x-2)>0
x^2-x-2=(x-1/2)²-9/4
The solution of inequality (x-a) / (x-a ^ 2) about X
Discussion on less than intermediate classification
It's mainly about the relationship between a and a ^ 2
When a = a ^ 2, that is, when a = 0 or a = 1, the left square can not be less than zero
When a < A ^ 2, that is, a > 1 or a < 0, a < x < A ^ 2
When a > A ^ 2, 0 < a < 1, a ^ 2 < x < a
Discussion by situation
1。 A
If the inequality x2 + A + 1 ≥ 0 holds for any real number 0 ＜ x ＜ 1 / 2, find the value range of constant a
Because 0 = - 1