It is known that when x = minus 3, the value of the fifth power of the polynomial ax plus the third power of BX plus CX minus 1 is 100. Find the value of this polynomial when x = 3

It is known that when x = minus 3, the value of the fifth power of the polynomial ax plus the third power of BX plus CX minus 1 is 100. Find the value of this polynomial when x = 3

a(-3)^5+b(-3)^3+c(-3)-1=100
Because it's all odd power, so
The original formula is - a × 3 ^ 5-b × 3 ^ 3-C × 3-1 = 100, that is - (a × 3 ^ 5 + B × 3 ^ 3 + C × 3) = 101
A × 3 ^ 5 + B × 3 ^ 3 + C × 3 = - 101, so a × 3 ^ 5 + B × 3 ^ 3 + C × 3-1 = - 102
So when x = 3, the value of the polynomial is - 102

If the sum of a polynomial and 3x ^ 2 + 2Y ^ 2 is x ^ 2 + XY-2 molecule 1y ^ 2, find the polynomial

X ^ 2 + XY-2 molecule 1y ^ 2 - (3x ^ 2 + 2Y ^ 2)
=X²+XY-1/2Y²-3X²-2Y²
=-2X²+XY-5/2Y²

If X1 and X2 are the two roots of the equation x2 + 2x-2007 = 0, try to find the following values: (1) X12 + X22; (2) 1x1 + 1x2; (3) (x1-5) (x2-5); (4) | x1-x2 |

This is the two roots of the equation x2 + 2x-2007 = 0, which is the two roots of the equation x2 + 2x-2007 = 0, and \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(2) 2 − 4x1 · x2 = (− 2) 2 − 4 × (− 2007) = 450 2．

X ^ 2 + 2x-2007 = 0 has two real roots, which are X1 and x2. Find the absolute value of x1-x2

X1+X2=-2,X1*X2=-2007,
|X1-X2|=√［(X1+X2)^2-4X1*X2］=√［4+4×2007］=4√502
∴X1-X2=±4√502

It is known that X1 and X2 are the two roots of the equation x & # 178; - 3x-1 = 0. By using the relationship between the root and the coefficient, the value of X1 & # 178; x2 & # 178; can be obtained

Let x 1 and x 2 be the roots of the equation x & # 178; - 3x-1 = 0
From x1 · x2 = - 1
We get X & # 178; 1. X & # 178; 2 = (- 1) &# 178; = 1

Given that X1 and X2 are the two roots of the equation 3x & # 178; - 2x-1 = 0, find the value of the following equation;
（1）（x1-x2）²； （2）(x1-2x2)(x2-2x1)

x1+x2=2/3
x1x2=-1/3
So (x1-x2) &;
=(x1+x2)-4x1x2
=4/9+4/3
=16/9
Original formula = 5x1x2-2x1 & # 178; - 2x2 & # 178;
=9x1x2-2(x1²+2x1x2+x2²)
=9x1x2-2(x1+x2)²
=-3-8/9
=-35/9

It is known that X1 and X2 are the two real roots of the equation 4kx & # 178 about X; minus 4kx plus K plus 1 equals zero,
(1) Whether there is a real number k, so that (2x1-x2) (x1-2x2) = - 1.5 holds. If there is, find out the value of K; if not, explain the reason;
(2) Find the integer value of the real number k of the position integer (x1 / x2) plus x2 / X1 minus 2

4kx & # 178; - 4kx + K + 1 = 0 from WIDA's theorem: X1 + x2 = - B / a = 1x1 · x2 = C / a = K + 11, assuming (2x1-x2) (x1-2x2) = - 1.5, then: 2 (x1) &# 178; + 2 (x2) &# 178; - 5x1 · x2 = - 1.5  2 ((x1 + x2) &# 178; - 2 x1 · x2) - 5x1 · x2 = - 1.52 (x1 + x2) &# 178; - 9x1 · x2 =

The quadratic form was transformed into standard form by formula method, f (x1, X2, x3) = x1x2-x1x3-3x2x3

f(x1,x2,x3)=x1x2-x1x3-3x2x3
=(y1+y2)(y1-y2)-(y1+y2)y3-3(y1-y2)y3
=y1^2 - 4y1y3 - y2^2 + 2y2y3
=(y1-2y3)^2 - y2^2 + 2y2y3 - 4y3^2
=(y1-2y3)^2 - (y2-y3)^2 - 3y3^2
=z1^2-z2^2-3z3^2

Hello teacher, f (x1, X2, X3, x4) = x1x2 + x2x3 + x3x4 + x4x1, use the formula method to change the quadratic form into the standard form, and find the reversible transformation?
Collar X1 = Y1 + Y2, X2 = y1-y2, X3 = Y3, X4 = Y4, and then substitute, but I don't know how to make the formula. Is there any step or method to master, please

It's a rule. If you don't have a square term, you can get the square term first. You've done this. First, deal with the items with Y1, Y1 ^ 2 - Y2 ^ 2 + Y1 * Y3 + Y1 * Y4 - Y2 * Y3 + Y2 * Y4 = (Y1 + 1 / 2y3 + 1 / 2y4) ^ 2 -- this includes all the items with Y1 - 1 / 4y3 ^ 2 - 1 / 4y4 ^ 2 - 1 / 2y3y4 --

Formula F (x1, X2, x3) = x1x2 + 2x1x3-4x2x3
Converting quadratic form to standard form

Let X1 = Y1 + Y2, X2 = y1-y2, X3 = Y3
Then f = (Y1) ^ 2 - (Y2) ^ 2 + 2y1y3 + 2y2y3-4y1y3 + 4y2y3
= (y1)^2-(y2)^2-2y1y3+6y2y3
= (y1-3y2-y3)^2-10(y2)^2-(y3)^2
= (z1)^2-10(z2)^2-(z3)^2
Where Z1 = y1-3y2-y3 = - X1 + 2x2-x3
z2=y2=(x1-x2)/2
z3=y3=x3.