Given that the polynomial x ^ 3-3x ^ 2 + 5x + a can be divided by x ^ 2-x + 3, the value of constant a can be obtained

Given that the polynomial x ^ 3-3x ^ 2 + 5x + a can be divided by x ^ 2-x + 3, the value of constant a can be obtained

x^3-3x^2+5x+a
=x(x^2-x+3)-2(x^2-x+3)+6+a
6+a=0
a=-6

If the third power of the polynomial 3x + the second power of MX + NX + 1 can be divisible by the second power of X + 1, and the quotient is 3x + 1, find the nth power of (- M)
Find the value of (- M) to the nth power

It is not necessary to divide by polynomials
3x^3+mx^2+nx+1/(x^2+1)=3x+1
After obtaining m and N, judge whether to choose or not
3x^3+mx^2+nx+1=3x^3+x^2+3x+1
So m = 1, n = 3
The nth power of (- M) = - 1

To make 3x3 + MX2 + NX + 42 divisible by x2-5x + 6, the value of M and n should be ()
A. m=8、n=17B. m=-8、n=17C. m=8、n=-17D. m=-8、n=-17

The quotient of ∵ 3x3 ∵ x2 = 3x, 42 ∵ 6 = 7, ∵ 3x3 + MX2 + NX + 42 divisible by x2-5x + 6 is (3x + 7) and ∵ x2-5x + 6 (3x + 7) = 3x3-8x2-17x + 42 ∵ M = - 8, n = - 17

It is known that polynomial x ^ 3-3x ^ 2 + 5x + a can be divided by polynomial X-2, and the value of constant a is
A.-6
B.30
C.2
D.-2

A:
X ^ 3-3x ^ 2 + 5x + a can be represented by polynomial X-2
Then X-2 = 0, that is, x = 2 is one of the solutions of the equation x ^ 3-3x ^ 2 + 5x + a = 0
Substitute:
8-12+10+a=0
The solution is: a = - 6
Option a

13x-7.5x = 18.7x3 to solve the equation

13X-7.5X=18.7X3
5.5x=56.1
x=56.1/5.5
x=10.2

It is known that the ellipse (the focus is on the x-axis) and the line x + Y-1 = 0 intersect at two points a and B, M is the midpoint of the line AB, and a direction vector of the line OM is
（2012,503）.
1： Second, if the minimum distance between a focal point of the ellipse and a point on the ellipse is 2-radical 3, and there is a point p rain A. B on the ellipse to form a triangle, try to find the maximum PAB area of the triangle

Let the coordinates of a and B be (x1, Y1), (X2, Y2). Then the slope k AB = (y2-y1) / (X2, - x1), and the coordinates of the midpoint m be (x1 + x2 / 2, Y1 + Y2 / 2). Because the point is on the ellipse, let the standard equation of the ellipse be x / A + Y / b = 1.1. So X1 & # 178 / / A & # 178; + Y1 & # 178 / / B & # 178; = 1, X2 & #

What is the decimal number 25 in 8421BCD code?

The BCD code of two 8421 chips is 0010 in high order and 0101 in low order

（1/2+1/3+1/4+1/5+1/6+.+1/2013）*（1+1/2+..+1/2012）-（1+1/2+1/3+1/4+...+1/2013）*（1/2+.+1/2012）
（1/2+1/3+1/4+1/5+1/6+...+1/2013）*（1+1/2+1/3+1/4...+1/2012）-（1+1/2+1/3+1/4+...+1/2013）*（1/2+1/3+1/4+...+1/2012）

Let 1 / 2 + 1 / 3 + 1 / 4 +... + 1 / 2012 = a
Then the above formula
=(A+1/2013)*(A+1)-(1+A+1/2013)*A
After decomposition
=A²+(1+1/2013)A+1/2013-A²-(1+1/2013)A
=1/2013

Definition: if the function y = f (x) exists x0 (a < x0 < b) in a given interval [a, b] in the domain of definition, satisfying f (x0) = f (b) − f (a) B − a, then the function y = f (x) is called the "mean function" on [a, b], and x0 is its mean point. For example, y = X4 is the mean function on [- 1, 1], and 0 is its mean point. (1) judge whether the function f (x) = - x2 + 4x is on the interval [0, 9] Is it an average function? If so, find out its mean value; if not, explain the reason; (2) if the function f (x) = - x2 + MX + 1 is the mean value function on the interval [- 1,1], try to determine the value range of real number M

(1) From the definition, when the equation - x2 + 4x = f (9) − f (0) 9 − 0 of X has a real root in (0, 9), the function f (x) = - x2 + 4x is an average function in the interval [0, 9]. By solving - X2 + 4x = f (9) − f (0) 9 − 0 {x2-4x-5 = 0, we can get x = 5, x = - 1

Given that m and N are two of the equations x2-2003x + 2004 = 0, then the product of (n2-2004n + 2005) and (m2-2004m + 2005) is______ ．

∵ m and N are two of the equations x2-2003x + 2004 = 0, ∵ m2-2003m + 2004 = 0, n2-2003n + 2004 = 0, namely m2-2003m = - 2004, n2-2003n = - 2004, ∵ original formula = (- N + 1) (- M + 1) = Mn - (M + n) + 1, ∵ m + n = 2003, Mn = 2004, ∵ original formula = 2004-2003 + 1 = 2