A few factorized 7th grade math problems high score!! ）

Garbage, copy it and say that you will be divided.

1)(x-1)^2-(y+2)^2=(x+y-1+2)(x-1-y-2)=(x+y+1)(x-y-3)

2)(y^2-1)(y^2+8y+15)+15=y^4+8y^3+14y^2-8y=y(y^3+8y^2+14y-8)

3) When n 2+7n-n 2+5n-6=6(2n-1)n is a positive integer, 2n-1 is a positive integer, and 6(2n-1) is a multiple of 6.

So the value of the algebraic equation n(n+7)-(n-3)(m-2) is divisible by 6.

1) Pending coefficient method.

2) Substitution method.

3) Multiply first, and finally get 12n-6

6（n-1）

N is a positive integer, so it is divisible.

1、(a+b)(4-2a)

m(x-y)-n(y-x)=3m(x-y)+n(x-y)=(x-y)（3m-n）

3、a(a-b-c)+2b(b+c-a)-c(c-a+b) =a(a-b-c)-2b(a-b-c)+c(a-b-c)=(a-2b+c)(a-b-c)

In the end, do the math yourself, I hope you understand.

(1) (z squared - x squared - y squared) perfectly squared - 4 (x squared) (y squared).

x-y-z) (x+y-z) (x-y+z) (x+y+z) (2) 1-(a) - (b) + (a) (b) - 4ab

-1-a-b+ab)(-1+a+b+ab)(3)(x4 power) + (x3 power)-3 (x) -4x-4=(-2+x)(2+x)(1+x+x 2).

1. It is known that a b=8, and a+b=-6, the values of a and b are ( ).

One is -2 and one is -4The answer above... Obviously, a and b are equivalent, and it is not possible to determine who is -2 and who is -4

It is easier to understand by using Veda's theorem in reverse. If you haven't learned it, then use the elimination method!

2. Decomposition factor: x -5x-6 = (

x-6）.（x+1）

3. When k=7, the factor: x +kx+10=(

x+2)(x+5)

4. When k=-11, decompose the factor x +kx+10=(

x-1)(x-10)

5. If x + px + 8 = (x-2) (x-q), then p= , q=

p=-6 ，q=4

6. Decomposition factor: a to the fourth power - 5a +4 (writing process).

Let a be squared x. The original form is a univariate quadratic polynomial.

After decomposition, it is ((x-1)(x-4)).Just swap the x back.

6. Decomposition factor: x -x-2y-4y (writing process).

The method is called Group Decomposition. Just take the recipe of X and Y and you're done.

x-1/2）（x-1/2）+（2y+1/4）（2y+1/4）

Sorry, but the computer level doesn't work, it won't be squared. So let's use the above method to represent the square. Forgive me.

7. Decomposition factor: 4x -4xy+y -44x+22y+40 (writing process).

The method is called the undetermined coefficient method (mainly because of the occurrence of terms such as 4xy. The method of the previous question will not work)

The coefficients of the higher terms 4x and y can be decomposed into (2x+by+c)(2x+1 b+d) (all coefficients to be determined).

Set the formula above. Since he and the original formula are the same formula, the corresponding coefficients are equal. Get the equation for BCD.

Solve the equation to obtain; b = -1, c and d are -2 and -20

So the answer is (2x-y-2) (2x-y-20).

8. Decomposition factor: (1-x) (1-y) + 4xy (writing process).

1-x²-y²+x²y²+4xy=（-x²-y²+2xy）（1+2xy+x²y²）

x-y）²+1+xy）²=1+xy+x-y）（1+xy-x+y）

1. It is known that a b=8, and a+b=-6, the values of a and b are (-2) (-4).

2. Decomposition factor: x -5x-6 = (x + 1) (x - 6).

3. When k=7, the factor: x +kx+10=(x+2)(x+5).

4. When k=-11, decompose the factor x +kx+10=(x-1)(x-10).

5. If x + px + 8 = (x-2) (x-q), then p = -2 and q = 4

6. Decomposition factor: a to the fourth power - 5a +4 (writing process).

Solution: Original formula = a 4 - a 2-4a 2 + 4 = a 2 (a 2-1) -4 (a 2-1).

a^2-1)(a^2-4)

a-1)(a+1)(a+2)(a-2)

6. Decomposition factor: x -x-2y-4y (writing process).

Solution: Original formula = (x 2-4y 2) - (x + 2y) = (x - 2y) (x + 2y) - (x + 2y).

x+2y)(x-2y-1)

7. Decomposition factor: 4x -4xy+y -44x+22y+40 (writing process).

Solution: Original = (4x 2-4xy+y 2)-22(2x-y)+40=(2x-y) 2-22(2x-y)+40

2x-y-2)(2x-y-20)

8. Decomposition factor: (1-x) (1-y) + 4xy (writing process).

Solution: Original formula = 1-x 2-y 2+x 2y 2+4xy=x 2y 2+2xy+1-(x 2+y 2-2xy).

xy+1)^2-(x-y)^2

xy+1-x+y)(xy+1+x-y)

1. ab=8 a+b=-6

Solution: (a+b)2=a2+2ab+b2

36=a2+b2+16

a2+b2=20

I'll do that. Excuse me.

Answer: a=-4 b=-2

1, -4, -2 or -2, -4

2、（x-6)(x+1)

3、(x+5)(x+2)

4、(x-1)(x-11)

6、(a2-)(a2-1)=(a+2)(a-2)(a-1)(a+1)7、（2x-y)2-22(2x-y)+40=(2x-y-2)(2x-y-20)

x2-y2+x2y2+4xy=(1+2xy+x2y2)-(x2-2xy+y2)=(1+xy)2-(x-y)2=(1+xy-x+y)(1+xy+x-y)

The 2 after the letter and parentheses is an exponent, copy it yourself and take a look.

Original = 4x y z (a-b) (6yz-5x+2x y z) I can only turn to this hidden cavity.

Original stove regret = - (5x+3y).

Original = - (2z-1).

Original = Original = (999 2 3 + 1000 3) (999 2-311000 3).

Primitive = 105 -2 (100 + 5) (100-5) + 95 Primitive = a +(2b) = a+2b) (a -ab + b ) primitive = [(x+y)-(x-y)] x+y) +x+y)(x-y)+(x-y)].

2y×(3x²+y²)

2a²+4a)-(a²-4)

a²+4a+4

a+2) 0 is equal if and only if the circle guesses a=-2.

A is a positive number. (2a²+4a)-(a²-4)>02a²+4a＞a²-4

Original = 9x (x -8y).

9x²(x-2y)(x²+2xy+4y²)a-b=2x²+2x+1

x²+2x+1+x²

x+1)²+x² ≥0a≥b

Can you write it down on paper and take a picture of it, so I can't read it.

==。=2[（x²+y²）²4xy（x²+y²）+4x²y²]=2（x²+y²-2xy）²

2(x-y) 4( denoted as (x-y) to the power of 4).

4。From (a +b) (a + b -8) +16 = 0, (a + b ) 8 (a + b ) + 16 = 0, (a + b -4) = 0, a + b -4 = 0, obtain: a + b = 4.

(1) Decomposition factor: (x+y) (x-y)+(x+y)(x-y).

x+y)²(x-y)+(x+y)(x-y)²

(x+y)(x-y)][x+y)+(x-y)] extract (x+y)(x-y)]

2x(x+y)(x-y)

2) Calculated using the squared difference formula: 2005 -2000

4005x5

Reference A -b = (a+b)(a-b).

3) Calculated with the formula of square difference: 100 -99 +98 -97 +96 -95 +....2²-1²

100+99)(100-99)+(98+97)(98-97)+.2+1)(2-1) (the difference between the subtractions is 1).

100+99+98+97+..2+1 (1 to 100 plus bar).

Reference A -b = (a+b)(a-b) 1+2+3+.n=(1+n)n/2

4) Decomposition factor: (x+y) (x-y)-(x+y)(x-y).

x+y)³（x-y)-（x+y)(x-y)³

(x+y)(x-y)][x+y)+(x-y)] extract (x+y)(x-y)]

x+y)(x-y)[(x+y+x-y)(x+y-x+y)] put the latter with this a -b = (a+b)(a-b))).

x+y)(x-y)(2x*2y)

4xy(x+y)(x-y)

1 (x+y) (x-y)+(x+y)(x-y) x+y)(x-y)(x+y)+(x+y)(x-y)(x-y)(x-y) Extract the common factor (x+y)(x-y).

x+y)(x-y)(x+y+x-y)

2x(x+y)(x-y)

4 (x+y) (x-y)-(x+y)(x-y) =(x+y) (x+y)(x-y)-(x+y)(x-y)(x-y) Extract the common factor (x+y)(x-y).

x+y)(x-y)[(x+y)²-x-y)²]=(x+y)(x-y)[(x+y+x-y)(x+y-x+y)=(x+y)(x-y)(2x)(2y)

4xy(x+y)(x-y)