How to do X & # 178; + 3A & # 178; = 4ax-2a + 1? x²+3a²=4ax-2a+1~

How to do X & # 178; + 3A & # 178; = 4ax-2a + 1? x²+3a²=4ax-2a+1~

x²+3a²=4ax-2a+1
x²-4ax+4a²=a²-2a+1
(x-2a)²=(a-1)²
The results are as follows
x-2a=±(a-1)
That is: x = 3a-1 or x = a + 1
x²+3a²=4ax-2a+1
x^2-4ax+4a^2=a^2-2a+1
(x-2a)^2=(a-1)^2
x-2a=±(a-1)
x=2a±(a-1)
x=3a-1,x=a+1
First, it can be changed into a general formula by formula or factorization
a+1，3a-1
Set a = {x | X & # 178; + (2a-3) x-3a = 0, X ∈ r}, B = {x | X & # 178; + (A-3) x + A & # 178; - 3A = 0, X ∈ r},
Let a ≠ B, and a ∩ B ≠ &;, denote a ∪ B with examples
A∩B≠Ф
So the two equations have common roots
Let the common root be B
Then B ^ 2 + (2a-3) B-3A = 0
b^2+(a-3)b+a^2-3a=0
subtract
(2a-3-a+3)b-3a-a^2+3a=0
ab-a^2=0
a(b-a)=0
If a = 0, then both equations are x ^ 2-3x = 0, which contradicts a ≠ B
So a ≠ 0
So B = a
So the common root is a
Substituting a ^ 2 + a (2a-3) - 3A = 0
3a^2-3a-3a=0
a^2-2a=0
a=0,a=2
We have just obtained a ≠ 0, so a = 2
So a x ^ 2 + X-6 = 0
(x+3)(x-2)=0
x=2,x=-3
B x^2-x-2=0
(x-2)(x+1)=0
x=2,x=-1
So a ∪ B = {2, - 3, - 1}
1, given the complete set u = {1,2,3,4}, a = {x | & # 144; X & # 178; - 5x + M = 0, X belongs to u}, find the complement of a in U, M
2, given the complete set u = R, set a = {x | - 2 less than x less than or equal to 5}, B = {x | - M + 1 less than or equal to x less than or equal to 2m-1}, and B is included in a, find the value range of real number M
3. Given the set a = {x | X & # 178; + 4x = 0} B = {x | X & # 178; + ax + a = 0}, if B is contained in a, find the condition that a satisfies
1. Let two of the equations X & # 178; - 5x + M = 0 be x1, X2 ᙽ X1 + x2 = 5,1 + 4 = 5,2 + 3 = 5. The equation x & # 178; - 5x + M = 0 can be (x-1) (x-4) = 0 or (X-2) (x-3) = 0, that is, X & # 178; - 5x + 4 = 0 or X & # 178; - 5x + 6 = 0. When a = {1,4}, M = 4, the complement of a in U is {2,3} when a = {2,3}, M = 6, the complement of a in U
ab（a+b）² -（a+b）-1
a²+ab-2b²
x²-4xy+4y²-3x+3y+2
ab（a+b）² -（a+b）-1
There is something wrong with the question
a²+ab-2b²
=(a+2b)(a-b)
x²-4xy+4y²-3x+3y+2
There is something wrong with the question
ab（a+b）² -（a+b）+1
=(a²+ab-1)(ab+b²-1)
a²+ab-2b²
=(a+2b)(a-b)
x²-4xy+4y²-3x+6y+2
=(x-2y)-3(x-2y)+2
=(x-2y-2)(x-2y-1)
I'm sorry, I changed 1 and 3. I think you copied the wrong question
a²+ab-2b²=(a2-b2)+(ab-b2)
=(a+b)(a-b)+b(a-b)
=(a-b)(a+2b)
ab（a+b）² -（a+b）-1
a²+ab-2b²
=(a-b)(a+2b)
x²-4xy+4y²-3x+3y+2
=(x-2y)²-3(x-y)+2
=(x-2y-2)(x-2y+2)
Simplify the following formulas (a > 0, b > 0) [(2a) 1 / 2 power + (3b) - 1 / 4 power] [2A 1 / 2 power - (3b) - 1 / 4 power]
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[(2a) ^ (1 / 2) + 3B ^ (- 1 / 4)] [(2a) ^ (1 / 2) - 3B ^ (- 1 / 4)] = (2a) ^ [(1 / 2) * 2] - 3B ^ [(- 1 / 4) * 2] = 2a-1 / (radical 3b)
Square difference formula
2 (square of 2A + a) - 11 (square of 2A + a) - 6
Classmate: ha ha, do you find that 2A ^ 2 + a can be regarded as an X first
Then there is 2 * x ^ 2-11x-6 = (X-6) (2x + 1) {cross method}
Substituting x = 2A ^ 2 + a into (2a ^ 2 + a-6)*
(4a ^ 2 + 2A + 1) = (a + 2) * (2a-3) * (4a ^ 2 + 2A + 1) {cross method}
Let (2a square + a) = a
So the original formula is the square of 2a-11a-6
According to factorization
So the original formula = (a-6) (2a + 1)
And a = the square of 2A + a
So the original formula = (a-6) (2a + 1) = (square of 2A + a-6) (square of 2A + A + 1)
And then according to the factorization
So the original formula = (A-2) (2a + 3) (the square of 2A + A + 1)
2 (square of 2A + a) - 11 (square of 2A + a) - 6
=[2 (square of 2A + a) + 1] (square of 2A + a-6) = (square of 4A + 2A + 1) (2a-3) (a + 2)
2 (square of 2A + a) - 11 (square of 2A + a) - 6
=[2 (square of 2A + a) + 1] (square of 2A + a-6) = (square of 4A + 2A + 1) (2a-3) (a + 2)
(1) It is known that a and B satisfy A2 + B2 + 4a-8b + 20 = 0, try to decompose (x2 + Y2) - (B + ax); (2) calculate: (1-122) (1-132) (1-142); (3) calculate: (1-122) (1-132) (1-142); (3) calculate: (1-122) (1-132) (1-142); (3) calculate: (1-122) (1-132) (1- (1-120082) (1-120092); (3) let a = 1999x + 1998, B = 1999x + 1999, C = 1999x + 2000, find the value of A2 + B2 + C2 AB AC BC
(1) A2 + B2 + 4a-8b + 20 = 0, (a + 2) 2 + (B-4) 2 = 0, so a = - 2, B = 4, (x2 + Y2) - (4-2xy) = x2 + Y2 + 2xy-4 = (x + y) 2-4 = (x + y + 2) (x + Y-2); (2) original formula = (1-12) × (1 + 12) × (1-13) × (1 + 13) × (1 + 14) × (1 + 14) × ×（1-12008）×（1+12008）×（1-12009）×（1+12009）=12×32×23×43×34×… (3) 2 (A2 + B2 + C2 AB AC BC) = (a-b) 2 + (A-C) 2 + (B-C) 2; when a = 1999x + 1998, B = 1999x + 1999, C = 1999x + 2000, (a-b) 2 + (A-C) 2 + (B-C) 2 = (- 1) 2 + (- 2) 2 + (- 1) 2 = 1 + 4 + 1 = 6
Factorization:
1.x^2+2x-1
2.x^2+6x+6
3.4x^2+12x+7
4. Let a ^ 2, B ^ 2, C ^ 2, d ^ 2 be integers, and satisfy (AB + CD) ^ 2 + (AD BC) ^ 2 = 36, find a ^ 2 + B ^ 2 + C ^ 2 + D ^ 2
5. Given that X and y are real numbers, find the minimum value of 5x ^ 2-4xy + 4Y ^ 2 + 12x + 25
1、x^2+2x-1=(x+1)²-2=（x+1+√2)(x+1-√2）
2、x^2+6x+6=(x+3)²-3=（x+3+√3)(x+3-√3）
3、4x^2+12x+7=(2x+3)²-2=（2x+3+√2)(2x+3-√2）
4、(ab+cd)^2+(ad-bc)^2=36
（a²+c²)(b²+d²）=36
Because a & sup2;, C & sup2;, B & sup2;, D & sup2; are integers
So a & sup2; + C & sup2; and B & sup2; + D & sup2; are integers, so there are
1）a²+c²+b²+d²=0+1+6+6=13
2）a²+c²+b²+d²=2+2+3+3=10
5、5x^2-4xy+4y^2+12x+25
=x²-4xy+4y²+(4x²+12x+9)+16
=(x-2y)²+(2x+3)²+16
Because X and y are real numbers, both (x-2y) & sup2; and (2x + 3) & sup2; are greater than or equal to 0,
So when (x-2y) & sup2; and (2x + 3) & sup2; are both 0, the original formula has a minimum,
That is, when x = - 3 / 2, y = - 3 / 4, the original formula has a minimum value of 16
1.x^2+2x-1=x^2+2x+1-2=(x+1)²-2=（x+1+√2）（x+1-√2）
2. Similarly, x ^ 2 + 6x + 6 = x ^ 2 + 6x + 9-3 = (x + 3) & sup2; - 3 = (x + 3 + √ 3) (x + 3 - √ 3)
3.4x^2+12x+7=4x^2+12x+9-7=(2x+3)²-2=（2x+3+√2）（2x+3-√2）
4. Expansion: A & sup2; B & sup2; +... Expansion
1.x^2+2x-1=x^2+2x+1-2=(x+1)²-2=（x+1+√2）（x+1-√2）
2. Similarly, x ^ 2 + 6x + 6 = x ^ 2 + 6x + 9-3 = (x + 3) & sup2; - 3 = (x + 3 + √ 3) (x + 3 - √ 3)
3.4x^2+12x+7=4x^2+12x+9-7=(2x+3)²-2=（2x+3+√2）（2x+3-√2）
4. Expand: A & sup2; B & sup2; + C & sup2; D & sup2; + A & sup2; D & sup2; + B & sup2; C & sup2; = 36 (A & sup2; + C & sup2;) (B & sup2; + D & sup2;) = 36
5.x²-4xy+4y²+(4x²+12x+9)+16=(x-2y)²+(2x+3)²+16
So when x = - 3 / 2, y = x / 2 = - 3 / 4, the minimum of the original formula is 16
1、(x+1)²-2=（x+1+√2）（x+1-√2）
2、(x+3)²-3=（x+3+√3）（x+3-√3）
3、(2x+3)²-2=（2x+3+√2）（2x+3-√2）
4. Expansion: A & sup2; B & sup2; + C & sup2; D & sup2; + A & sup2; D & sup2; + B & sup2; C & sup2; = 36 (A & sup2; + C & sup2;)... Expansion
1、(x+1)²-2=（x+1+√2）（x+1-√2）
2、(x+3)²-3=（x+3+√3）（x+3-√3）
3、(2x+3)²-2=（2x+3+√2）（2x+3-√2）
4. Expansion: A & sup2; B & sup2; + C & sup2; D & sup2; + A & sup2; D & sup2; + B & sup2; C & sup2; = 36 (A & sup2; + C & sup2;) (B & sup2; + D & sup2;) = 36
5、x²-4xy+4y²+(4x²+12x+9)+16=(x-2y)²+(2x+3)²+16
So when x = - 3 / 2, y = x / 2 = - 3 / 4, the minimum of the original formula is 16
You said the list can be broken down. But most of them have roots. First, let the enumerated formula equal to zero, then find the root, and then decompose the formula to (x-a) (X-B), a and B are the roots.
The fourth question, first decompose, and then get (A2 + C2) (B2 + D2) = 36. It can be discussed in categories, 1 * 36 = 36, 2 * 18 = 36, 3 * 12 = 36, and so on. So A2 + B2 + C2 + D2 = 37 or 20 or 15 or 13 or 12
The fifth question. Take y or X as a constant, and use the method of finding the root, that is, which method of the previous questions to factorize. But... Unfold
You said the list can be broken down. But most of them have roots. First, let the enumerated formula equal to zero, then find the root, and then decompose the formula to (x-a) (X-B), a and B are the roots.
The fourth question, first decompose, and then get (A2 + C2) (B2 + D2) = 36. It can be discussed in categories, 1 * 36 = 36, 2 * 18 = 36, 3 * 12 = 36, and so on. So A2 + B2 + C2 + D2 = 37 or 20 or 15 or 13 or 12
The fifth question. Take y or X as a constant, and use the method of finding the root, that is, which method of the previous questions to factorize. Then make the brackets equal to zero. 5x2 - (4y-12) x + 4y2 + 25 = 0. Don't be afraid of trouble. And then the minimum is, you do it yourself. Put it away
If a ^ 2 + B ^ 2-4a-8b + 20 = 0, find B ^ (- a)
That is, (A & # 178; - 4A + 4) + (b-8b + 16) = 0
(a-2)²+(b-4)²=0
So A-2 = B-4 = 0
a=2,b=4
So the original formula = 4 ^ (- 2) = 1 / 16
1.（x+1）（x+2）-x-10
2.（x2-2x（x2-2x-4）+4
3.x2-3xy-18y2-3x-9y
2.(x2-2x）（x2-2x-4）+4
1.（x+1）（x+2）-x-10
=x^2+2x+x+2-x-10
=x^2+2x-8
=(x-2)(x+4)
2.（x2-2x（x2-2x-4）+4
You write the title clearly
3.x2-3xy-18y2-3x-9y
=x^2-3xy-18y^2-3(x+3y)
=(x-6y)(x+3y)-3(x+3y)
(x+3y)(x-6y-3)
1.（x+1）（x+2）-x-10=x^2+3x+2-x-10=x^2+2x-8=(x-2)(x+4)
2.（x^2-2x)（x^2-2x-4）+4=(x^2-2x)^2-4(x^2-2x)+4=(x^2-2x-2)^2
3. x^2-3xy-18y^2-3x-9y=(x-6y)(x+3y)-3(x+3y)=(x-6y-3)(x+3y)
（x+1）(x+2) - x - 10
= xx + 3x + 2 - x - 10
= xx + 2x - 8
= (x + 4) ( x - 2 )
（x^2-2x)（x^2-2x-4）+4=(x^2-2x)^2-4(x^2-2x)+4=(x^2-2x-2)^2
x2-3xy-18y2-3x-9y
=x^2-3xy-18y^2-3(x+3y)
=(x-6y)(x+3y)-3(x+3y)
(x+3y)(x-6y-3)
1.（x+1）（x+2）-x-10
=x^2+2x+x+2-x-10
=x^2+2x-8
=(x-2)(x+4)
2.（x2-2x（x2-2x-4）+4
The title is not clear, the brackets are not right...
3.x2-3xy-18y2-3x-9y
=x^2-3xy-18y^2-3(x+3y)
=(x-6y)(x+3y)-3(x+3y)
(x+3y)(x-6y-3)
1、（x-2)(x+4)