If (root 3x + 2) + (2-y) 2 = 0, then the value of x 2 is

If (root 3x + 2) + (2-y) 2 = 0, then the value of x 2 is

(root 3x + 2) + (2-y) 2 = 0
So 3x + 2 = 0,2-y = 0
So x = - 2 / 3, y = 2
So x 2 = 4 / 9

It is known that x = 4- 3, find x4-6x3-2x2 + 18x + 23 X2-8x + 15

If (x-4) 2 = 3, that is, x2-8x + 13 = 0, then x2-8x = - 13
Molecular x4-6x3-2x2 + 18x + 23,
=x4-8x3+2x3-2x2+18x+23，
=x2（x2-8x）+2x3-2x2+18x+23，
=-13x2+2x3-2x2+18x+23，
=2x3-16x2+x2+18x+23，
=2x（x2-8x）+x2+18x+23，
=-26x+x2+18x+23，
=x2-8x+23，
=-13+23，
=10，
The denominator is x2-8x + 15 = - 13 + 15 = 2,
∴x4-6x3-2x2+18x+23
x2-8x+15=10
2=5．
So the answer is: 5

The maximum value of the quadratic function y = f (x) = ax ^ 2 + 2aX + 1 in the interval [- 3,2] is 4. Find the value of A

I finally saw a classmate with a little level. Kid, if you can't do this kind of question, it's excusable to run to ask. You're better than those who don't know the basic knowledge of series. You're much better than the people who come to ask Baidu
In this paper, we investigate the maximum value problem of quadratic function on closed interval
Since it is known to be a quadratic function, there must be a ≠ 0. The axis of symmetry is x = - 1
(next, I'm going to start the classification discussion. It's not easy to draw pictures on it. You have to watch my process and draw on the paper yourself.)
The axis of symmetry is inside [- 3,2], and the midpoint of the interval is - 0.5, so the axis of symmetry is close to the left end point - 3
① When a > 0, the opening is upward, and the maximum value is obtained at the end point,
Since the axis of symmetry is close to the left end point, the maximum value is f (2) = 8A + 1
Let 8A + 1 = 4, and a = 3 / 8
② When a < 0, the opening is downward, and the maximum value is obtained at the symmetry axis, so the maximum value is f (- 1) = 1-A
Let 1-A = 4 and a = - 3
To sum up, a = 3 / 8 or - 3
Note: the problem of the value range of quadratic function on closed interval, in the final analysis, is to discuss the "position relationship between the axis of symmetry and the interval", and of course, combined with the opening

If the minimum value of the quadratic function y = AX2 + 4x + A-1 is 2, then the value of a is () A. 4 B. 3 C. -1 D、 4 or -1

∵ the quadratic function y = ax2-4x + A-1 has a minimum value of 2,
∴a＞0，
Y min = 4ac − B2
4a=4a(a−1)−(−4)2
4a=-13a2-4=-17，
A = - 1 or 4,
∵a＞0，
∴a=4．
Therefore, a

Find the value range of the quadratic function y = - 2x square + 6x on the following definition domain (what is the definition domain, how to find it?) (1) The domain {x belongs to Z | 0 ≤ x ≤ 3}; (2) The domain is defined as [- 2,1] In addition, what is the difference between definition domain and value domain?

The definition domain is the value range of X and the value range of Y. the drawing of the value range of quadratic function can find out the opening of the original function through the vertex coordinates and the definition domain, and the symmetry axis is x = 3 / 2 vertex coordinates (3 / 2, 9 / 2), and the X axis intersects with (0,0) (3,0) to draw an image, you can see that (1) y belongs to Z and 0 ≤ y ≤ 9 / 2 (2) function in [-...]

It is known that the negative integer solution of the inequality (X-2) / x < (1 + 2x) / 3-1 is the solution of equation (2x-1) / 3 - (a + x) / 2 = 1. Find the value of A

Solving inequality:
(x-2)/x0,
So x > 0
Inequality (X-2) / x0}
There is no negative integer solution
So the equation (2x-1) / 3 - (a + x) / 2 = 1 has no solution,
solve equations:
(2x-1)/3-(a+x)/2=1,
4x-2-3x-3a=6,
x=8+3a.
If the equation (X-2) / X

If inequality system x＞2m+1 Then x is the solution of M ＞ A. -1 B. -3 C. - 1 or - 3 D. -1＜m＜1

According to the meaning of the question, when 2m + 1 = - 1, the solution is m = - 1, while m + 2 = - 1 + 2 = 1,
When m + 2 = - 1, M = - 3 and 2m + 1 = - 5 ＜ - 1,
So when m = - 3, the inequality system
x＞2m+1
The solution of X ＞ m + 2 is x ＞ - 1
Therefore, B

If 3x? - X-1 + 0, find the value of 6x cubic power + 7x? - 5x + 2001 Thank you. It's urgent

6x^3+7x^2-5x+2001
=(2x+3)*(3x^2-x-1)+2004
=(2x+3)*0+2004=2004

The expansion polynomial f (x) = x ^ 4-5x ^ 3 + x ^ 2-3x + 4 according to the power of (x-4)

Expand f (x) = x ^ 4-5x ^ 3 + x ^ 2-3x + 4 according to the power of x-4: calculate the derivatives of each order first
f'(x)=4x^3-15x^2+2x-3.
f''(x)=12x^2-30x+2.
f'''(x)=24x-30
f''''(x)=24.
F '' '"' (x) = 0 (thus, after expansion, the remainder is 0, that is, this is an error free expansion.)
The following data are obtained: F (4) = - 56, f '(4) = 21, f' '(4) = 74, f' '' (4) = 66, f '' '(4) = 24
So f (x) = x ^ 4-5x ^ 3 + x ^ 2-3x + 4
=-56+21(x-4)+(74/2!)(x-4)^2+(66/3!)(x-4)^3+(24/4!)(x-4)^4
=-56+21(x-4)+37(x-4)^2+11(x-4)^3+(x-4)^4

The 2nd power of 5x + the 2nd power of 4-3x - the 2nd power of 5x-2x - 5 + 6x, where x = - 3

The 2nd power of 5x + the 2nd power of 4-3x - the 2nd power of 5x-2x - 5 + 6x
=（5x²-3x²-2x²）+(6x-5x)+(4-5)
=x-1
=-3-1
=-4