On the polynomial of X, the quadratic term coefficient, the first order coefficient and the constant term of the square + 2x of - 3x are ()

On the polynomial of X, the quadratic term coefficient, the first order coefficient and the constant term of the square + 2x of - 3x are ()

On the polynomial of X, the coefficients of quadratic term, primary term and constant term of square + 2x of - 3x are (- 3 and 2 and 0), respectively

Let f (x) = 2cos square x + root 3 times sin2x + a (a belongs to R, a is a constant). 1. Find the increasing region of F (x) if x belongs to R 2. When x belongs to the 0-2 fraction of the closed interval, the maximum value of F (x) is 4, then what is the value of a? 3. Under the condition of 2, find the set of X which satisfies f (x) = 1 and X belongs to [- π, π]

1. F (x) = 2cos square x + root 3 times sin2x + a = 1 + cos2x + 3sin2x + a = 2Sin (2x + π / 6) + 1 + a
Let - π / 2 + 2K π

Given the square of a + ax + X-1, B = 3x of the third power - 2x + = 1 (a is a constant) 1, a, B does not contain the square term of X, find a? 2, simplify b-2a on the above

A = the square of AX + X-1, B = 3x to the third power - 2x + 1 (a is a constant)
It does not contain x 2
∴a=0
B-2a = (3x to the third power - 2x + 1) - 2 (x-1)
=3x³-2x+1-2x+2
=3x³-4x+3

If the positive integer x satisfies the inequality 3x + 4 ≥ 5x + 2 and satisfies the equation 2 (x + a) - 4A + = 0, try to find the square root of A Junior high school eight mathematics on the problem, urgent

2> X = 2x, X is a positive integer, so x = 1. Put it into the equation to find a. what is the last + sign in your equation?

(1) - (x? - 2x) + (- 3x + 2x? (2) - 2 (a? - 3AB) - 3 (ab-2a?) Reduction and re evaluation: (1) 9x + 6x? - 3 (x - two thirds x? 2) where x = - 1 (2) One half (2x? - 6x-4) - 4 (- 1 + X + quarter x? 2) where x = 5

1) (x + 2x + 2x + 2x + 2x + 2x + 2x + 2x + 2x (2x + 2x + 2x + 2x + 2x (2x + 2x + 2x + 2x + 2x + 2x + 2x + 2x + 2x + 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x / 2x-6x-4) - 4 (2x-6x-4) - 4-4 (2x when x = 5, the original formula = - 35 + 2 = - 33

Who can solve the two calculation problems: 3x? - 1 = 8 (2x-1) 2 = 0.008

3x²-1=8
3x²=9
x²=3
x=±√ 3
（2x-1)²=0.008
（2x-1)²=1/125
2x-1=±1/25√5
2x=1±1/25√5
x=1/2±1/50√5

24[1/8(3x-1+x²)-2/3(x³-2x²+3)-x}

24[1/8(3x-1+x²)-2/3(x³-2x²+3)-x]=24*1/8(3x-1+x²)-24*2/3(x³-2x²+3)-24*x=3(3x-1+x²)-16(x³-2x²+3)-24x=9x-3+3x²-16x³+32x²-48-24x=-16x³+3x&#...

How to match 2x? 3x + 3 to 2 (x-3 / 4) 2 + 15 / 8 Detailed steps are required

Original formula =2 (x 2 -3x/2) +3
=2(x²-3x/2+9/16-9/16)+3
=2(x²-3x/2+9/16)-9/8+3
=2(x-3/4)²+15/8

The distance from the circle x2 + Y2 + 2x + 4y-3 = 0 to the straight line x + y + 1 = 0 is Points of 2 share () A. 1 B. 2 C. Three D. Four

By transforming the equation of circle into standard equation, we get: (x + 1) 2 + (y + 2) 2 = 8,
The coordinates of the center of the circle are (- 1, - 2), and the radius is 2
2，
The distance from the center of a circle to the straight line x + y + 1 = 0 d = 2
2=
2，
Then the distance from the circle to the straight line x + y + 1 = 0 is
There are three points of 2
Therefore, C is selected

A value of C whose distance from the line 2x + y + C = 0 is equal to 1 is () A. 2 B. Five C. 3 D. 3 Five

The equation of the circle can be reduced to: (x-1) 2 + (y + 2) 2 = 4, so the center of the circle m (1, - 2), radius r = 2, it is easy to know if and only if the distance D ∈ (1, 3) from m to the line L: 2x + y + C = 0, the distance between two points on ⊙ m and the line L is equal to 1, and from D = | C | 5 ∈ (1,3), C ∈ (− 35, − is obtained