Factorization Hurry, Hurry, Hurry!

Because a 2 + b 2 + c 2 = ab + ac + bc, 2a 2 + 2b 2 + 2c 2 = 2ab + 2ac + 2bc i.e.: 2a 2 + 2b 2 + 2c 2-2ab-2ac -2bc 0a 2-2ab + b 2) + (a 2-2ac + 2c 2) + (b 2-2bc + c 2) + 0

a-b) 2+(a-c) 2+(b-c) 2 0 so, a-b) 2 0

a-c）^2＝0

b-c）^2＝0

So, a b c

So, the triangle ABC is an equilateral triangle.

Solution: Multiply both sides of the equation by 2 at the same time.

2a 2+2b 2+2c 2=2ab+2ac+2bc move the algebraic test on the left to the right.

2a 2+2b 2+2c 2-2ab-2ac-2bc 0 is obtained by group decomposition method.

a^2-2ab+b^2）+（a^2-2ac+c^2）+（b^2-2bc+c^2）＝0

It is obtained using the perfect square formula.

a-b) 2+(a-c) 2+(b-c) 2 0 because the square of any number can only be zero or positive.

So the above equation is 0 0 0 0

Both a-b 0, a-c 0, b-c 0

So a=b, a=c, b=c

Since a=b=c, this triangle is equilateral.

Multiply both sides of the equation by 2.

2a 2+2b 2+2c 2=2ab+2ac+2bc, then move the right item to the left and assign it.

a^2+b^2-2ab+a^2+c^2-2ac+b^2+c^2-2bc=0

Collated (a-b) 2+(a-c) 2+(b-c) 2=0, from this we get a=b=c, so the triangle is an equilateral triangle.

= 8,x1+x2 = m, so the eggplant quietly bears fruit with such as possible as Pei Qi next few:

x1 = 1,x2 = 8,m=-9;

x1 = 2,x2 = 4,m=-6;

x1 =-1,x2 =-8,m= 9;

x1 =-2,x2 =-4,m= 6;

2.Original = a 4-13a 2-48

Question 1: m can be 6 -6 9 -9, that is, decompose 8 into 2 and 4 multiplied by -2 and -4, and there are 1 and 8, and -1 and -8

1 2 The two questions are all decomposed by the cross multiplication method to balance the deficit of the empty mountain.

x*x*x(x+1)(x-1) xy(xxyy-1) squared ( xy-5 ) xy+2 )

x+y)squared-1 = (x+y-1)(x+y+1)= (

A square + b square - 2ab) - (6a-6b) + 9 = (a-b) square - 6 (a-b) + 9

a-b-3).

: x to the fifth power - x cubic = x 3 (x 2-1) = x 3 (x + 1) (x - 1).

x^5·y^5-2y^3·x^3+xy=xy(x^4y^4-2x^2y^2+1)=xy(x^2y^2-1)^2=xy(xy+1)^2(xy-1)^2

x square·y squared - 3xy-10 = (xy-5) (xy + 2).

x square + y square - 1-2xy = (x-y) 2-1 = (x-y + 1) (x-y-1).

Calculations with factorization:

Knowing a=10000, b=9999, find (a-squared+b-squared-2ab)-(6a-6b)+9

A square + b squared - 2ab) - (6a-6b) + 9 = (a-b) 2-6 (a-b) = 1-6 = -5

-4x(x -4x+6)=-4x[(x -4x+4)+2]=-4x(x-2+2i)(x-2-2i) (Note: If you have not learned complex numbers, the answer is the polynomial before the first equal sign).

8a(a-b)²-12(b-a)³=8a(a-b)²+12(a-b)³=4(a-b)²[2a+3(a-b)]=4(a-b)²(5a-3b)

2a^（m+1）+4a^m-2a^(m-1) =2a^(m-1)(a²+2a-1)=2a^(m-1)[(a+1)²-2]=2a^(m-1)(a+1+√2)(a+1-√2)

2a²b²-4ab+2=2[(ab)²-2ab+1]=2(ab-1)²

x²+y²)²4x²y²=(x²)²2x²x²+(y²)²4x²y²=(x²)²2x²x²+(y²)²=(x²-y²)²

Alas, the holiday is too long, I can't write a word, I play too much game, and I can't see the question 1Original = -4x (x -4x+6) However: x -4x+6 in =16-24=-8, that is, there is no solution.

2. 4（a-b）²（5a-3b）

3. 2a^(m-1)（a²+a-1）4. 2（ab-1）²

5. (x²-y²)²

6. (x+y-2)²

It's just my own answer, not necessarily right, check it out for yourself. I haven't touched a book for two months, but I miss my passion for mathematics before graduation.

It's good to give you tips, the first one proposes a 4x, the second one puts forward the negative sign first, the third, the matching type, and the fourth one regards ab as a whole, and the child should use his own brain for homework, otherwise how to become smart.

Original = (a+1)[(a+1)-2b]+b =(a+1) He Ji-2(a+1)*b+b =[a+1)-b] =a-b+1)

Lookout pants.