(1) Solving the system of inequalities 2-5x

(1) Solving the system of inequalities 2-5x

(1) Left half: 2-5x < x + 3 = = > - 5x-x < 3-2 = 1 = = > - 6x < 1 = = = > x > - 1 / 6 (1) Right half: x + 3 < 3x / 2 = = > (3x / 2) - x > 3 = = > x / 2 > 3 = = = > x > 6 ...
Let f (x) = determinant, what is the coefficient of x ^ 3?
Line by line, the main diagonal is the fourth power of X, the others are either the first power of X or the zero power of X
The coefficient of x ^ 3 is 0
The determinants 1, 2, 3, 3, x 2, 4, 5 2 are used to find the coefficients of X
I know how to calculate x, but how can X be determined
1 2 3
3 x 2
4 5 2
The coefficient of X is M22=
13
42
= 2 - 12
= -10
Solution of 3x ^ 2 + 2 √ 2x-5 = 0 formula
You can use the root formula to answer
X=[-b±(b^2-4ac)^(1/2)]/2a
According to the title: X1 = (√ 17 - √ 2) / 3
X2=-(√17+√2)/3
If the solution of inequality (3a-b) x-4a-3b > 0 is x < 2 / 3, then the solution of AX + b > 0 is?
(3a-b)x-4a-3b＞0
(3a-b)x>4a+3b
X-B
Then x
What is the slope formula
Let a (x1, Y1), B (X2, Y2), x1 ≠ x2
Then the slope of AB is k = (y2-y1) / (x2-x1)
Tan + angle
Slope k = (y2-y1) / (x2-x1)
How to solve the problem of 4x ^ 3-3x ^ 2-1 = 0
4x^3-3x^2-1=0
It can be seen that x = 1 is a root, denoted as x1
(4x^3-3x^2-1)/(x-1)=4x^2+x+1
4x^2+x+1=0 ==>
X2 = (- 1 + radical (16-4)) / 8 = 0.308
X3 = (- 1 + radical (16-4)) / 8) = -0.558
x1=1,x2=0.308,x3=-0.558
4x³-3x²-1=(3x³-3x²)+(x³-1)=3x²(x-1)+(x-1)(x²+x+1)
=(x-1)(4x²+x+1)=(x-1)[(2x+1/4)²+15/16]=0
So x = 1
4x³-3x²-1=(3x³-3x²)+(x³-1)=3x²(x-1)+(x-1)(x²+x+1)
=(x-1)(4x²+x+1)
4X & sup2; + X + 1 is always greater than 0
x-1=0
X=1
It is known that the image of the first-order function y = 4x-4 intersects with the X axis at point a, and the image of the positive scale function y = 2x intersects with point B. the coordinates of points a and B are obtained
Intersection with X axis at point a
A（1,0）
The solution is 2x-y = 4x-4
B（2,4）
Take y = 0 into x = 1, that is, a coordinate is (1,0). Because the images of y = 4x-4 and y = 2x intersect at point B, X and y are equal, so the solution of 2x = 4x-4 is x = 2. Take x = 2 into the original analytical formula to get y = 4
Given the set a = {y | y = log2x, X ＞ 1}, B = {y | y = (12) x, X ＞ 1}, find (1) a ∩ B; (2) (CRA) ∪ B
(1) ∫ set a = {y | y = log2x, X ＞ 1} = {y | y ＞ 0}, B = {y | y = (12) x, X ＞ 1} = {y | 0 ＜ y ＜ 12}, a ∩ B = {y | 0 ＜ y ＜ 12}. (2) from (1), we know that CRA = {y | y ≤ 0}, B = {y | y ＜ 12}, and (CRA) ∪ B = {y | y ＜ 12}
Solve the following inequality 5x + 125 ≤ 0.6-2x ＞ 0-x + 1 ＞ 7x-32 (1-3x) ＞ 3x + 20, 2x-1 of 2 ＜ X of 2, 1-2x of 3 ≥ 4-3x of 6
5x + 125 ≤ 0.6-2x ＞ 0-x + 1 ＞ 7x-32 (1-3x) ＞ 3x + 20 2 / 2 2x-1 ＜ 2 / 3 1-2x ≥ 6 / 4-3x
5X+125≤0==>xx xxxx
5X+125≤0 6-2X＞0 -X+1＞7X-3 2（1-3X）＞3X+20
x≤25 x