Using the collocation method to solve the equation x & # 178; - 6x + 3 = 0

Using the collocation method to solve the equation x & # 178; - 6x + 3 = 0

x²-6x+3=0
x²-6x+9=6
（x-3）²=6
x-3=±√6
x=3±√6
The correct way to arrange the polynomial - A2 + a3 + 1-A by the ascending power of the letter A is ()
A. a3-a2-a+1B. -a-a2+a3+1C. 1+a3-a2-aD. 1-a-a2+a3
∵ in the polynomial - A2 + a3 + 1-A, the exponent of - A is 1, the exponent of - A2 is 2, the exponent of A3 is 3, and the ascending order of a is 1-a-a2 + a3
The circled number represents the power. For example, 2 (2) represents the second power of 2
1.（6x-1）（2x-1）+（3x-1）（x-1）+ x ②
2.x ⑨ +x ⑥ +x ③ -3
3.6x ④ +7x ③ -36x ② -7x + 6
4.a③b - ab③ + a② +b② +1
5.3x② - 11xy + 6y② - xz - 4yz -2z②
6.x② - 8xy + 15y② +2x - 4y -3
If you can't answer all of them, please answer as many as you can,
Or give an idea, such as which can be made by formula or cross, I can't do it any more, just ask for help
X ^ 2 is the square, in turn, x ^ 3 is the cubic power 1.12x ^ 2-6x-2x + 1 + 3x ^ 2-3x-x + 1 + x ^ 2 = 16x ^ 2-12x + 2 = (2x ^ 2-1) (8x ^ 2-2) = 2 (x ^ 2-1) (2x ^ 2-1) 2
Using the collocation method to solve the equation: X & # 178; - 6x + Y & # 178; + 10Y = - 34
x²-6x+9+y²+10y+25=0
(x-3)²+(y+5)²=0
∴x-3=0
y+5=0
∴x=3
y=-5
It is necessary to arrange the polynomial x2y-2x3y2-3 + 4xy3 in ascending order of the letter X______ .
The terms of the polynomial are X2Y, - 2x3y2, - 3, 4xy3, arranged by the ascending power of the letter X is - 3 + 4xy3 + x2y-2x3y2. So the answer is - 3 + 4xy3 + x2y-2x3y2
-X to the third power Z + X to the fourth power y
3x（a-b）+2y（b-a）
(2a+b)(2a-3b)+(2a+5b)(2a+b)
-X to the third power Z + X to the fourth power y
=x³(xy-z)
3x（a-b）+2y（b-a）
=3x(a-b)-2y(a-b)
=(a-b)(3x-2y)
(2a+b)(2a-3b)+(2a+5b)(2a+b)
=(2a+b)[(2a-3b)+(2a+5b)]
=(2a+b)(4a+2b)
=2(2a+b)²
The equation is solved by collocation method: 6x2-x-12 = 0
The original equation can be reduced to x2-16x = 2, ∧ x2-16x + (112) 2 = 2 + (112) 2, the formula is (x-112) 2 = 289144, ∧ x-112 = ± 1712, the solution is X1 = 32, X2 = - 43
Remove brackets: the square of the polynomial 3A minus the cube of a minus 1 minus 1 minus the square of AB is arranged by the ascending power of the letter A
The original formula = 3A & # 178; - A & # 179; - 1-1 + AB & # 178;
=-2+ab²+3a²-a³
3a²-a³-1-(1-ab)²
=3a²-a³-1-(1-2ab+a²b²)
=-2+2ba+(3-b²)a²-a³
=-The cube of 2-AB + 3a-a
1. Calculate the value of the following equation:
[（1^4＋¼）（3^4＋¼）…… （19^4＋¼）]÷[（2^4＋¼）（4^4＋¼）…… （20^4＋¼）]
2. Known multiplication formula: A ^ 5 + B ^ 5 ≈ (a + b) (a ^ 4-A ^ 3B + A ^ 2B ^ 2-AB ^ 3 + B ^ 4)
a^5－b^5≈（a－b）（a^4＋a^3b＋a^2b^2＋ab^3＋b^4）
With or without the above formula, factorize x ^ 8 + x ^ 6 + x ^ 4 + x ^ 2 + 1
The approximate equal sign of the two formulas should be equal sign. Wrong number.
no one shows any interest in? One of them is OK.
（2）(x^8+x^6+x^4+x^2+1)
=(x^2-1)(x^8+x^6+x^4+x^2+1)/(x^2-1)
=(x^10-1)/(x^2-1)
=(x^5+1)(x^5-1)/(x^2-1)
=(x+1)(x^4+x^3-x+1)(x-1)(x^4+x^3+x^2+x+1)/(x^2-1)
=(x^4+x^3-x+1)(x^4+x^3+x^2+x+1)
To solve the binary linear equation: X (X-2) - 3x & # 178; = 0
x(x-2)-3x²=0
x(x-2-3x)=0
x(-2x-2)=0
X = 0 or x = - 1