The inequality of solution about X: 2x2 + kx-k ≤ 0

The inequality of solution about X: 2x2 + kx-k ≤ 0

From 2x2 + kx-k = 0, we can get △ = K2 + 8K, let △ = 0, we can get k = 0 or - 8. ① when △ < 0, i.e. - 8 < K < 0, the solution set of the original inequality is ∈. ② when △ = 0, i.e., K = 0 or - 8, the solution set of the original inequality is {0} or {2}. ③ when △ > 0, i.e., k > 0 or K < 8, from 2x2 + kx-k = 0, we can get x = − K ± K2 + 8k4. The solution set of the original inequality is {x | K − K2 + 8k4 ≤ x ≤− K + K2 + 8k4}
The inequality kx2-2x + 6K ＜ 0 (K ≠ 0) about X is known. If the solution set of the inequality is {x | x ≠ 1K}, find the value of the real number K
Because K ≠ 0, it can be regarded as a quadratic function, y = kx2-2x + 6K. When k > 0, the opening is upward, and the quadratic function is less than 0, the scope of X will always be limited, but it can't; when k < 0 is, the opening is downward, ∵ kx2-2x + 6K < 0, ∧ Δ = 4-24k2 ≤ 0, the solution set of∧ inequality is {x | x ≠ 1K}, ∧ Δ = 0, the solution set of k = - 66, so the value of K is: - 66
Given that the solution set of inequality KX ^ 2-2x + 6K < 0 about X is r, find the value range of K
The solution set of inequality KX ^ 2-2x + 6K < 0 is r
When k = 0, - 2x
Let y = KX Λ 2-2x + 6K
① If k = 0 - 2x ＜ 0, then x ＞ 0 does not hold
② If it is a quadratic function, y < 0, X ∈ R, then the image is open phase, so K < 0
y=k（x∧2-2x/k+6）=k［（x-1/k）∧2-1/k∧2+6］
So x = 1 / K has a maximum
Ymax = K (6-1 / K Λ 2) = 6k-1 / K < 0
Simplify K Λ 2 ＞ 1 / 6... Expansion
Let y = KX Λ 2-2x + 6K
① If k = 0 - 2x ＜ 0, then x ＞ 0 does not hold
② If it is a quadratic function, y < 0, X ∈ R, then the image is open phase, so K < 0
y=k（x∧2-2x/k+6）=k［（x-1/k）∧2-1/k∧2+6］
So x = 1 / K has a maximum
Ymax = K (6-1 / K Λ 2) = 6k-1 / K < 0
Simplify K ∧ 2 ＞ 1 / 6 - √ 6 / 6 ＜ K ＜√ 6 / 6
Because K ＜ 0, so - √ 6 / 6 ＜ K ＜ 0
To sum up, {K | - √ 6 / 6 ＜ K ＜ 0} ask: - 2x ＜ 0, isn't x definitely greater than 0?
A mathematical problem, known about X inequality KX ^ 2-2x + 6k0)
(1) If the solution set of the inequality is {x | 2}
1.2 and 3 are the two equations KX ^ 2-2x + 6K = 0, one of which can be taken in to get k = 0.4
2. F (x) = KX ^ 2-2x + 6K: F (2)
1 k=0.4
2 K is greater than 1 / / 6
Square root of 3 1 / 6
If the image of function f (x) = loga (4 + ax) and the image of function g (x) = Log1 / a (a + x) (a > 0 and a ≠ 1) are symmetric with respect to the line y = B (B is a constant)
Find a + B
Because of symmetry, then f (x) + G (x) = 2B holds for any x ∈ D (D is the intersection of two function domains). The comparison coefficient shows that a = 2, B = 0
Factorization x ^ 4 + 2x ^ 3Y + 3x ^ 2Y ^ 2 + 2XY ^ 3 + y ^ 4
=(x^2+x+1)^2
When deducing the formula of circle area, divide the circle into several parts and put it together into an approximate rectangle. It is known that the length of the rectangle is 6.42 cm more than the width, so the area of the circle can be calculated
The formula already knows why the step of 42 (3.14-1) = 3 (CM) should be reduced by one
The length is half of the circumference π R
Width is the radius r of a circle
πr-r=6.42
R = 6.42 ÷ (3.14-1) = 3cm
3 × 3 × 3.14 = 28.26 square centimeter
Because the sum of the length of a rectangle is the circumference of a circle, and the width of a rectangle is the radius of a circle.
So: let the radius of the circle be r cm, and the length of the rectangle be 3.14 × 2R △ 2 = 3.14r
Width is r, so 3.14r-r = 6.42
(3.14-1)r=6.42
r=6.42÷（3.14-1）
R=3
That's how it works, okay? Why 3. 14r-r = 6.42
(3.14-1) r = 6.4
Because the sum of the length of a rectangle is the circumference of a circle, and the width of a rectangle is the radius of a circle.
So: let the radius of the circle be r cm, and the length of the rectangle be 3.14 × 2R △ 2 = 3.14r
Width is r, so 3.14r-r = 6.42
(3.14-1)r=6.42
r=6.42÷（3.14-1）
R=3
That's how it works, okay? Question: why 3. What about 14 minus one for radius
Calculate the following questions: 2.5Y + 4Y 8x + X 7n-n
2.5y+4y=6.5y 8x+x=9x 7n-n=6n
The value of solution X and y of equation 3x-2y = | a | also satisfies | 2x + Y-1 | + (x-3y) ^ 2 = 0, and | a | + a = 0
^2 is square, the faster the better, the better today
|2x+y-1|+(x-3y)^2=0
2x+y-1=0
x-3y=0
x=3/7
y=1/7
3x-2y=1=|a|
|a|+a=0
a=-|a|=-1
-1
Let a > 0. A ≠ 1 and f (x) = a ^ LG (x ^ 2-2x + 3) have the maximum value, the solution set of inequality a ^ (x ^ 2-5x + 7) > 1 is obtained
Let a > 0., and a ≠ 1, f (x) = a ^ LG (X & sup2; - 2x + 3) have the maximum,
∵ X & sup2; - 2x + 3 = (x-1) & sup2; + 2 is always positive and has a minimum value of 2,
To make the function f (x) = a ^ LG (X & sup2; - 2x + 3) have the maximum value,
Then there must be 0