On the equation 2x ^ 2-3x-2k = 0 of X, there is a real root in (- 1,1), so we can find the value range of K

On the equation 2x ^ 2-3x-2k = 0 of X, there is a real root in (- 1,1), so we can find the value range of K

9+4*2k*2>=0
That is k > = - 9 / 16
Obviously, when k = - 9 / 16, it is in line with the meaning of the question
Let f (x) = 2x ^ 2-3x-2k
F (1) f (- 1)
If the equation has a real root in (- 1,1), then the function f (x) = 2x ^ 2-3x-2k has an intersection with the x-axis in (- 1,1)
Then f (- 1) * f (1)
The root of equation [2x-3] [3x + √ 2] = 0 is
The root of equation [2x-3] [3x + √ 2] = 0 is
x1=1.5,x2=(-1/3)√2.
The two roots that make the equation 0 are:
X = 3 / 2 or x = - √ 2 / 3?
A fourth-order determinant, the first line is 1 1, the second line is 1 1-x 1 1, the third line is 1 12-x 1, and the fourth line is 1 13-x
What are the roots of an equation equal to 0,
D = all lines minus line 1
1 1 1 1
0 -x 0 0
0 0 1-x 0
0 0 0 2-x
= -x(1-x)(2-x)
All roots are 0,1,2
Given the inequality (4a-3b) about X, the solution set X of x greater than 2b-a is less than 4 / 9, find the solution set of ax greater than B
The solution set of (4a-3b) x ＞ 2b-a is x ＜ 4 / 9. It can be seen that the solution set of 4a-3b ＜ 0 (4a-3b) x ＞ 2b-a x ＜ (2b-a) / (4a-3b) = 4 / 9 is B / a = 5 / 6, that is, B = 5A / 6 is substituted by 4a-3b ＜ 0 to get 4a-35a / 6 ＜ 0 to get a ＜ 0 ax ＞ B x ＜ B / a = 5 / 6 ax ＞ B is x ＜ 5 / 6
Seeking various slope formulas in Mathematics
I don't know how to say it. It's the slope formula. I think the product of the slopes of two vertical lines is - 1
(1) If the slope of two lines exists and is not zero: L1: y = K1X + b1l2: y = k2x + B2, if L1 ⊥ L2, then K1 * K2 = - 1, if L1 / / L2, then K1 = K2 (2) for any two lines: L1: a1x + b1y + C1 = 0l2: a2x + b2y + C2 = 0, if two lines are vertical, then A1A2 * b1b2 = 0, if two lines are parallel, then A1B
Linear slope formula: k = (y2-y1) / (x2-x1) y = KX + B; (Y-Y1) / (y2-y1) = (x-x1) / (x2-x1); k = (y2-y1) / (x2-x1); Y / B + X / a = 1; y = - B / ax + B; k = TaNx ' Almost
Given x ^ 2 + y ^ 2-4x-2y + 5 = 0, find the value of [2 / 3x √ 9x + y ^ 2 √ [x / y ^ 2]]
After (X-2) ^ 2 + (Y-1) ^ 2 = 0
We get x = 2, y = 1
The following formula is not clear
Just substitute X and Y in
The sum of this function is determined by the function (- K + 2) of degree (- K + 1)
k = (6+2)/(1+3) = 2
b = 6-2 = 4
So: y = 2x + 4
Bring in a, B coordinates
-2=-3k+b
6=k+b
The solution is k = 2, B = 4
The function formula is 2x + 4
Solving inequality log2 (2x) × log2 (x / 4) < 4
Analysis:
According to the meaning of the title: x > 0
Then the original inequality can be reduced to: [(log2 2) + (log2 x)] * [(log2 x) - (log2 4)]
log2(2x)×log2(x/4)＜4
[log2(2)+log2(x)][log2(x)-log2(4)]
The solution set of (4a-3b) x > 2b-a is the solution set of XB
Let (4a-3b) x = 2b-a
x=（4a-3b)/(2b-a)
Let x = (4a-3b) / (2b-a)
Substituting (4a-3b) x > 2b-a
BX / 2 > 9b / 8
Because XB
The results show that (- B / 8) x ＞ B
So x ＞ - 8
The slope of straight line and its formula
1. If the direction vector of a straight line is a = (m, n) (M is not equal to 0), then the slope of the straight line is K=__________ .
2. When finding the slope of the tangent at the point (x0, Y0) on the curve, it can be transformed into a function, and K can be obtained by using the derivative knowledge=____________ .
Please fill in the blank on the line, 3Q!
(1)
k = tan a = n/m
(2)
k=(dx)/(dy)
lim y1-y0
x1->x0 k=-------
y1->y0 x1-x0
K = (y-y0) / (x-x0) this "similar to" straight line point oblique form
k=n/m
k=(y-y0)/(x-x0)