5x + 3x-20 equals 100 to find x

5x + 3x-20 equals 100 to find x

8x=120
x=15
4a2 + b2-4a + 10B + 26 = 0
4A * a + b * b-4a + 10B + 26 = 4 (a-0.5) (a-0.5) + (B + 5) (B + 5) = 0 because the square is greater than or equal to 0, so a-0.5 = 0, B + 5 = 0, so the positive solution is a = 0.5, B = - 5
I want factorization plus 50`
By the way, I still need answers`````
When (3x-42) is divided by 6 x, the result is 10
(3x - 42) ÷ 6 = 10
3x - 42 = 10 × 6
3x - 42 = 60
3x = 60 + 42
3x = 102
x = 102 ÷ 3
x = 34
4A * a + b * b-4a + 10B + 26 = 0, find the value of a and B
Hello, no passers-by
∵4a²＋b²－4a＋10b＋26＝0
∴（4a²－4a＋1）＋（b²＋10b＋25）＝0
∴（2a－1）²＋（b＋5）²＝0
2 A-1 = 0 and B + 5 = 0
∴a＝1/2,b＝－5
I need 300 simple factorization problems in grade two of junior high school. The more I do, the better I will add scores
^2 is the second power
1)x^2+2xy+y^2
2)x^2-y^2
1.a^4-4a+3
2.(a+x)^m+1*(b+x)^n-1-(a+x)^m*(b+x)^n
3.x^2+(a+1/a)xy+y^2
4.9a^2-4b^2+4bc-c^2
5.(c-a)^2-4(b-c)(a-b)
The original formula = a ^ 4-a-3a + 3 = (A-1) (a ^ 3 + A ^ 2 + A-3)
2.[1-(a+x)^m][(b+x)^n-1]
3.(ax+y)(1/ax+y)
4.9a^2-4b^2+4bc-c^2=(3a)^2-(4b^2-4bc+c^2)=(3a)^2-(2b-c)^2=(3a+2b-c)(3a-2b+c)
5.(c-a)^2-4(b-c)(a-b)
= (c-a)(c-a)-4(ab-b^2-ac+bc)
=c^2-2ac+a^2-4ab+4b^2+4ac-4bc
=c^2+a^2+4b^2-4ab+2ac-4bc
=(a-2b)^2+c^2-(2c)(a-2b)
=(a-2b-c)^2
1.x^2+2x-8
2.x^2+3x-10
3.x^2-x-20
4.x^2+x-6
5.2x^2+5x-3
6.6x^2+4x-2
7.x^2-2x-3
8.x^2+6x+8
9.x^2-x-12
10.x^2-7x+10
11.6x^2+x+2
12.4x^2+4x-3
To solve the equation: (x's square + 5x-6) 1 / 2 = (x's square + X + 6) 1 / 2
Although cross multiplication is difficult to learn, once we learn it and use it to solve problems, it will bring us a lot of convenience. Here are my personal opinions on cross multiplication
1. The method of cross multiplication: multiplication on the left side of cross equals the coefficient of quadratic term, multiplication on the right side equals the constant term, cross multiplication and addition equals the coefficient of primary term
2. The use of cross phase multiplication: (1) to decompose the factor by cross phase multiplication. (2) to solve quadratic equation with one variable by cross phase multiplication
3. Advantages of cross phase multiplication: the speed of solving problems by cross phase multiplication is relatively fast, which can save time, and the amount of calculation is not large, so it is not easy to make mistakes
4. The disadvantages of cross multiplication: 1. Some problems are easy to solve by cross multiplication, but not every problem is easy to solve by cross multiplication. 2. Cross multiplication is only suitable for quadratic trinomial problems. 3. Cross multiplication is difficult to learn
5. Examples of solving multiplication problems
1) Use cross multiplication to solve some simple and common problems
Example 1 decomposes M & sup2; + 4m-12 into factors
Analysis: in this problem, the constant term - 12 can be divided into - 1 × 12, - 2 × 6, - 3 × 4, - 4 × 3, - 6 × 2, - 12 × 1. When - 12 is divided into - 2 × 6, it is in line with this problem
Because 1-2
1 ╳ 6
So M & sup2; + 4m-12 = (m-2) (M + 6)
Example 2 decomposes 5x & sup2; + 6x-8 into factors
Analysis: in this problem, 5 can be divided into 1 × 5, - 8 can be divided into - 1 × 8, - 2 × 4, - 4 × 2, - 8 × 1
Because 1 2
5 ╳ -4
So 5x & sup2; + 6x-8 = (x + 2) (5x-4)
Example 3 solves the equation x & sup2; - 8x + 15 = 0
Analysis: if X & sup2; - 8x + 15 is regarded as a quadratic trinomial about X, then 15 can be divided into 1 × 15,3 × 5
Because 1-3
1 ╳ -5
So the original equation is deformable (x-3) (X-5) = 0
So X1 = 3, X2 = 5
Example 4. Solve the equation 6x & sup2; - 5x-25 = 0
Analysis: take 6x & sup2; - 5x-25 as a quadratic trinomial of X, then 6 can be divided into 1 × 6,2 × 3, - 25 can be divided into - 1 × 25, - 5 × 5, - 25 × 1
Because 2-5
3 ╳ 5
So the original equation can be changed to form (2x-5) (3x + 5) = 0
So X1 = 5 / 2, X2 = - 5 / 3
2) Use cross multiplication to solve some difficult problems
Example 5: Factoring 14x & sup2; - 67xy + 18y & sup2
Analysis: if 14x & sup2; - 67xy + 18y & sup2; is regarded as a quadratic trinomial about X, then 14 can be divided into 1 × 14,2 × 7,18y & sup2; and y.18y, 2y.9y, 3y.6y
Solution: because 2 - 9y
7 ╳ -2y
So 14x & sup2; - 67xy + 18y & sup2; = (2x-9y) (7x-2y)
Example 6 decomposes 10x & sup2; - 27xy-28y & sup2; - x + 25y-3 into factors
Analysis: in this problem, we should arrange this polynomial into the form of quadratic trinomial
Solution 1: 10x & sup2; - 27xy-28y & sup2; - x + 25y-3
=10x²-（27y+1）x -（28y²-25y+3） 4y -3
7y ╳ -1
=10x²-（27y+1）x -（4y-3）（7y -1）
=[2x -（7y -1）][5x +（4y -3）] 2 -（7y – 1）
5 ╳ 4y - 3
=（2x -7y +1）（5x +4y -3）
Note: in this problem, 28y & sup2; - 25y + 3 is decomposed into (4y-3) (7y-1) by cross phase multiplication, and then 10x & sup2; - (27y + 1) x - (4y-3) (7y-1) is decomposed into [2x - (7y-1)] [5x + (4y-3)]
Solution 2: 10x & sup2; - 27xy-28y & sup2; - x + 25y-3
=（2x -7y）（5x +4y）-（x -25y）- 3 2 -7y
=[（2x -7y）+1] [（5x -4y）-3] 5 ╳ 4y
=（2x -7y+1）（5x -4y -3） 2 x -7y 1
5 x - 4y ╳ -3
Note: in this problem, first, 10x & sup2; - 27xy-28y & sup2; is decomposed into (2x - 7Y) (5x + 4Y) by cross phase multiplication, and then (2x - 7Y) (5x + 4Y) - (x - 25y) - 3 is decomposed into [(2x - 7Y) + 1] [5x - 4Y) - 3] by cross phase multiplication
Example 7: the solution of X equation: X & sup2; - 3ax + 2A & sup2; – ab - B & sup2; = 0
Analysis: 2A & sup2; – ab-b & sup2; can be factorized by cross phase multiplication
x²- 3ax + 2a²–ab -b²=0
x²- 3ax +（2a²–ab - b²）=0
x²- 3ax +（2a+b）（a-b）=0 1 -b
2 ╳ +b
[x-（2a+b）][ x-（a-b）]=0 1 -（2a+b）
1 ╳ -（a-b）
So X1 = 2A + B, X2 = a-b
5-7(a+1)-6(a+1)^2
=-[6(a+1)^2+7(a+1)-5]
=-[2(a+1)-1][3(a+1)+5]
=-(2a+1)(3a+8);
-4x^3 +6x^2 -2x
=-2x(2x^2-3x+1)
=-2x(x-1)(2x-1);
6(y-z)^2 +13(z-y)+6
=6(z-y)^2+13(z-y)+6
=[2(z-y)+3][3(z-y)+2]
=(2z-2y+3)(3z-3y+2).
For example... X ^ 2 + 6x-7
Because the coefficient before the first power X is 6
So, we can think of 7-1 = 6
The constant term of this formula is - 7
So we think of - 7 as 7 * (- 1)
So we made a cross
x +7
x -1
From (x + 7) · (x-1)
The factorization is successful
3ab^2-9a^2b^2+6a^3b^2
=3ab^2(1-3a+2a^2)
=3ab^2(2a^2-3a+1)
=3ab^2(2a-1)(a-1)
5-7(a+1)-6(a+1)^2
=-[6(a+1)^2+7(a+1)-5]
=-[2(a+1)-1][3(a+1)+5]
=-(2a+1)(3a+8);
-4x^3 +6x^2 -2x
=-2x(2x^2-3x+1)
=-2x(x-1)(2x-1);
6(y-z)^2 +13(z-y)+6
=6(z-y)^2+13(z-y)+6
=[2(z-y)+3][3(z-y)+2]
=(2z-2y+3)(3z-3y+2).
For example... X ^ 2 + 6x-7
Because the coefficient before the first power X is 6
So, we can think of 7-1 = 6
The constant term of this formula is - 7
So we think of - 7 as 7 * (- 1)
So we made a cross
x +7
x -1
From (x + 7) · (x-1)
The factorization is successful
3ab^2-9a^2b^2+6a^3b^2
=3ab^2(1-3a+2a^2)
=3ab^2(2a^2-3a+1)
=3ab^2(2a-1)(a-1)
x^2+3x-40
=x^2+3x+2.25-42.25
=(x+1.5)^2-(6.5)^2
=(x+8)(x-5)．
(6) cross multiplication
There are two cases of this method
① Factorization of formulas of type X ^ 2 + (P + Q) x + PQ
The characteristics of this kind of quadratic trinomial are: the coefficient of the quadratic term is 1; the constant term is the product of two numbers; the coefficient of the primary term is the sum of two factors of the constant term. Therefore, we can directly divide some quadratic trinomial factors of which the coefficient of the quadratic term is 1 into x ^ 2 + (P + Q) x + PQ = (x + P) (x + Q)
② Factorization of formulas of type KX ^ 2 + MX + n
If k = AC, n = BD and AD + BC = m, then KX ^ 2 + MX + n = (AX + b) (Cx + D)
The figure is as follows:
A B
X
C D
For example: because
1 -3
X
72
-3 × 7 = - 21, 1 × 2 = 2, and 2-21 = - 19,
So 7x ^ 2-19x-6 = (7x + 2) (x-3)
Formula of cross multiplication: decomposition of head and tail, cross multiplication, summation
(3) group decomposition method
A simple way to learn how to solve the equation is to divide the knowledge into groups
There are four terms or more than four terms in the equation which can be divided into groups
For example:
ax+ay+bx+by
=a(x+y)+b(x+y)
=(a+b)(x+y)
We divide ax and ay into a group, BX and by into a group, and use the law of multiplicative distribution to match each other
Similarly, this problem can be done in the same way
ax+ay+bx+by
=x(a+b)+y(a+b)
=(a+b)(x+y)
Several examples:
1. 5ax+5bx+3ay+3by
Solution: = 5x (a + b) + 3Y (a + b)
=(5x+3y)(a+b)
Note: different coefficients can be decomposed into groups. As above, 5AX and 5bx can be regarded as a whole, 3ay and 3By can be regarded as a whole, which can be easily solved by using the law of multiplicative distribution
2. x3-x2+x-1
Solution: = (x3-x2) + (x-1)
=x2(x-1)+(x-1)
=(x-1)(x2+1)
Use the dichotomy method, put forward the common factor method, put forward X2, and then combine to solve easily
3. x2-x-y2-y
Solution: = (x2-y2) - (x + y)
=(x+y)(x-y)-(x+y)
=(x+y)(x-y+1)
The dichotomy method and formula method a 2-B 2 = (a + b) (a-b) are used
758²—258² =(758+258)(758-258)=1016*500=508000
Solving equation 1 / 3x & # 178; - x-4 = 0 by collocation method
1/3x²-x-4=0
Multiply both sides by 3 to get the following result:
x²-3x-12=0
(x-3/2)²-9/4-12=0
(x-3/2)²=57/4
x-3/2=±√57/2
x=(3±√57)/2
If 4a2 + 4A + b2-6b + 10 = 0, then B / A-A / b =?
4a2+4a+b2-6b+10=0,
(2a+1)^2+(b-3)^2=0
a=-1/2,b=3
b/a-a/b
=-6+1/6
=-35/6
(2a+1)²+(b-3)²=0
b=3，a=-0.5
So the original formula = - 6 + 1 / 6 = - 35 / 6
The results are as follows
4a^2+4a+1+b^2-6b+9=0
(2a+1)^2+(b-3)^2=0
So: a = - 1 / 2, B = 3
So: B / A-A / b = - 5 / 6
(a-b)^5+(b-c)^5+(c-a)^5
First floor! That's Square!
Sorry, ooo_ KKK, due to my miscalculation and not looking at your answer carefully, I mistakenly think your answer is wrong. Here I apologize
Here I would like to correct:
(a-b)^5+(b-c)^5+(c-a)^5 =5(a-b)(b-c)(c-a)(a²+b²+c²-ab-bc-ac)
(a+b+c)^5-a^5-b^5-c^5=5(a+b)(b+c)(c+a)(a²+b²+c²+ab+bc+ac)】
[now I have changed my answer. Please read it carefully]
You should know the factor theorem
If x = a, the polynomial B &; X ^ n + C &; X ^ (n-1) + D &; X ^ (n-2) + +If the value of MX + n is 0, then (x - a) is a factor of the polynomial
Of course, the formula with multiple letters also holds
I think you should also know the rotation of symmetric polynomials
For a polynomial containing multiple letters, arrange the letters in it in a certain order (generally in the order before and after the alphabet), change the first letter into the second letter, change the second letter into the third letter, and so on, and change the last letter into the first letter. If the obtained polynomial is the same as the original polynomial, then the polynomial is the same
Rotational symmetric polynomials
For example, (a-b) ^ 5 + (B-C) ^ 5 + (C-A) ^ 5 is changed into (B-C) ^ 5 + (C-A) ^ 5 in the order of a, B and C, which is still the same as the original polynomial, so (a-b) ^ 5 + (B-C) ^ 5 + (C-A) ^ 5 is a rotational symmetric polynomial
For a rotational symmetric polynomial, if it is known that it contains a factor of low degree, it must contain other factors of the same type
For example, decomposition factor: (a-b) ^ 5 + (B-C) ^ 5 + (C-A) ^ 5
Substituting a = B into 0 ^ 5 + (A-C) ^ 5 + (C-A) ^ 5 = 0
So A-B is a factor of (a-b) ^ 5 + (B-C) ^ 5 + (C-A) ^ 5
Because (a-b) ^ 5 + (B-C) ^ 5 + (C-A) ^ 5 is a rotational symmetric polynomial
So B-C, C-A are also factors of (a-b) ^ 5 + (B-C) ^ 5 + (C-A) ^ 5
Because (a-b) ^ 5 + (B-C) ^ 5 + (C-A) ^ 5 is a rotational symmetric polynomial
According to its characteristics, let (a-b) ^ 5 + (B-C) ^ 5 + (C-A) ^ 5 = (a-b) (B-C) (C-A) [K (A & sup2; + B & S]