Bivariate linear equation, addition subtraction elimination method { 2x-5y=-21 4x+3y=23} {4x+7y=-19 4x-5y=17} {3（x-1）=y+5 5（y-1）=3）x+5} Addition subtraction elimination method

Bivariate linear equation, addition subtraction elimination method { 2x-5y=-21 4x+3y=23} {4x+7y=-19 4x-5y=17} {3（x-1）=y+5 5（y-1）=3）x+5} Addition subtraction elimination method

2x-5y=-21 ①
4x+3y=23 ②
① X2 is 4x-10y = - 42 ③
①-③
4x-10y-（4x+3y)=-42-23
y=5
Take y = 5 into 1
Find x = 2
4x+7y=-19 ①
4x-5y=17 ②
①-②
7y+5y=-19-17
y=-3③
Bring (3) into (1)
x=0.5
{3（x-1）=y+5
5（y-1）=3）x+5
3x-3=y+5① 3x-y=8 ③
5y-5=3x+5② 3x-5y=-10④
③-④
4y=18
y=9/2
Bring in ①
X = 25 / 6

The most difficult problem in the method of addition, subtraction and elimination of quadratic equation of two variables
It's all fractions
Choose ten or five
Please hurry up!

Who said there was no problem? Look, I gave you this
1、839x+727y=929
5/7x+83/8=42
2、78.3356x+57.242y-8.7678x=5.299765
8493.2843y-2854.73690x=-10058
3、9483x=8354x-4930y
39.93928x-849.38928x+389.4829y=-184.84930
4、282/997x+29923/34927y=1382.17398729
493x+3989y=0.01
5、2938x+4482y=0
2939/3298x+283+23982x-289y=1
Who can solve these problems in 12 hours without a computer or calculator is the best!

The problem of addition, subtraction and elimination of quadratic equation of two variables
Given x + 3Y = 5, then the value of the algebraic expression X-Y is
3x+y=-1,②

②×3 9x+3y=-3 ③
③-① 8x=-8
x=-1
Substituting x = - 1 into (1) y = 2
∴x=-1
y=2
The original formula = - 1-2 = - 3

Factorization 1 + x ^ 15
1+x^15

x^15+1
=(x^5)^3+1^3
=(x^5+1)(x^10-x^5+1)
=(x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+x+1)(x^10-x^5+1)
=[x^4(x+1)-x^3(x+1)+x^2(x+1)-x(x+1)+(x+1)](x^10-x^5+1)
=(x+1)(x^4-x^3+x^2-x+1)(x^10-x^5+1)

Decomposition factor: 1.7 (A-1) + X (a + 1)
1.7(a-1)+x(a+1)
Square of 2.3 (a-b) + 6 (B-A)
3.2 (m-n) - M (m-n)
4. Square of X (X-Y) - square of Y (Y-X)
5. M (square of a + square of B) + n (square of a + square of B)
6.18 (a-b) - the square of 12B (B-A)
7.（2a+b)(2a-3b）-3a(2a+b)
8. The square of X (x + y) (X-Y) - x (x + y)
There are three lawns in a university. The area of the first lawn is (a + b) square meters, the area of the second lawn is (a + b) square meters, and the area of the third lawn is (a + b) square meters. The total area of this lawn can be calculated

（a+b）（a+b）+a（a+b）+（a+b）b

Decomposition factor: (x2 + X + 1) (x2 + X + 2) - 12

Let x2 + x = y, then the original formula = (y + 1) (y + 2) - 12 = Y2 + 3y-10 = (Y-2) (y + 5) = (x2 + X-2) (x2 + X + 5) = (x-1) (x + 2) (x2 + X + 5). It shows that x2 + X + 1 can also be regarded as a whole. For example, if x2 + X + 1 = u, the same result can be obtained. Students who are interested may have a try. So the answer is (x-1) (x + 2) (x2 + X + 5)

Factorization factor X ^ 4 + x ^ 3 + x ^ 2 + X + 1
This is the original topic of our lecture. Later, we found that there was a problem
1. According to the teacher's method, decompose (undetermined coefficient: set as (x ^ 2 + ax + b) (x ^ 2 + CX + D)), and the answer is
(x ^ 2 + (1 + √ 5) / 2x + 1) (x ^ 2 + (1 - √ 5) / 2x + 1), but I have repeatedly verified that it is wrong, so is this solution wrong, or is this problem unable to factorize? (x ^ 2 represents the square of X, and √ is the root sign)
Master help, good Bonus!
So how to decompose x ^ 8 + x ^ 6 + x ^ 4 + x ^ 2 + 1? The answer is: (x ^ 4-x ^ 3 + x ^ 2-x + 1) (x ^ 4 + x ^ 3 + ^ x2 + X + 1). Is the answer wrong?

I'm afraid I'm a junior high school student. I haven't studied complex numbers. I have a look at it. This decomposition is right. There may be something wrong with the checking process. At the same time, I add that the quartic polynomial with real coefficients can be decomposed into the product of two quadratic expressions with real coefficients

How do X (x-1) (X-2) (x-3)... (x-n) decompose factors

This multiplies a polynomial of degree n, which requires n-order derivatives. The final result is that all of them will get 0. Only the term of the highest degree still exists after the n-order derivatives. The result of all n-order derivatives is n

Factorization of (a + 3) (A-7) + 24

(a+3)(a-7)+24
=a²-4a-21+24
=a²-4a+3
=(a-1)(a-3)

This is a factorization problem: (a + 1) (a +) (a + 2) (a + 3) + 1

(a+1)(a+2)(a+3)(a+4)+1
=(a+1)(a+4)(a+2)(a+3)+1
=(a^2+5a+4)(a^2+5a+6)+1
=(a^2+5a)^2+10(a^2+5a)+25
=(a^2+5a+5)^2