100 factoring problems in Mathematics in grade two of junior high school Give me a few simple factoring questions. If you can, write down the answers

100 factoring problems in Mathematics in grade two of junior high school Give me a few simple factoring questions. If you can, write down the answers

1. Decompose the following formulas into (1) 12a3b2-9a2b + 3AB; (2) a (x + y) - (a-b) (x + y); (3) 121x2-144y2; (4) 4 (a-b) 2 - (X-Y) 2; (5) (X-2) 2 + 10 (X-2) + 25; (6) A3 (x + y) 2-4a3c2.2. Calculate (1) 6... (1) 6
When x is the value, the sum of formula (4x-3) - 5 and 3x-1 is equal to 7
4x/3-5+(3x-1)=7
(4/3+3)x=7+5+1
13/3x=13
x=13÷13/3
X=3
The original equation is 4x / 3-5 + 3x-1 = 7
4x-15+9x=24
13x=39
X=3
Three
Factorization of the square of a-2a + b-2b + 2Ab + 1
a²-2a+b²-2b+2ab+1
=(a²+2ab+b²)-(2a+2b）+1
=(a+b)²-2(a+b)+1
=(a+b-1)²
Original formula = A & # 178; + 2Ab + B & # 178; - 2 (a + b) + 1
=（a+b）²-2（a+b）+1
=（a+b-1）²
36（x+2y）^2-25（x-2y）^2
36 (x + 2Y) & sup2; - 25 (x-2y) & sup2; let x + 2Y = m, x-2y = n, the original formula = 36m & sup2; - 25N & sup2; = (6m) & sup2; - (5N) & sup2; = (6m + 5N) (6m-5n) = [6 (x + 2Y) + 5 (x-2y)]] = (6x + 12Y + 5x-1
36（x+2y）^2-25（x-2y）^2
=[6(x+2y)+5(x-2y)][6(x+2y)-5(x-2y)]
=(11x+2y)(x+22y)
36（x+2y）²－25（x－2y）²
=[6（x+2y]²－[5（x－2y）]²
=[6（x+2y）＋5（x－2y）][6（x+2y）－5（x－2y）]
=（6x+12y+5x－10y)(6x+12y－5x+10y）
=（11x+2y)(x+22y)
36（x+2y）^2-25（x-2y）^2
=[6(x+2y）]²-[5(x-2y）]²
=(6x+12y+5x-10y)(6x+12y-5x+10y)
=(11x+2y)(x+22y)
36（x+2y）^2-25（x-2y）^2
=（6x+12y+5x-10y)(6x+12y-5x+10y)
=(11x+2y)(x+22y)
Original formula = (6 (x + 2Y) + 5 (x-2y)) (6 (x + 2Y) - 5 (x-2y))
=(6X+12Y+5X-10Y)(6X+12Y-5X+10Y)
=(11X+2Y)(X+22Y)
When what is the value of X, the sum of 43x-5 and 3x + 1 is equal to 9?
From the meaning of the question: 43x-5 + 3x + 1 = 9, to the denominator: 4x-15 + 9x + 3 = 27, to shift and merge: 13X = 39, coefficient into 1: x = 3
-Factorization of 8A ^ 2B + 2A ^ 3 + 8ab ^ 2
=a（-8ab+2a^2+8b^2）
=2a（a^2-4ab+4b^2）
=2a（a-2b）^2
Factorization exercise (it's better to write a simple process)
2(x-1)^2-x+1
-a^4+2a^2b^2-b^4
(x+2y)^2-4(x-3y)^2
9(2x-y)^2-6(2x-y)+1
(m^2-6m)^2+18(m^2-6m+81)
-12a(a-b)^3+6a^2(b-a)^3
2(x－1)²－x＋1＝2(x－1)²－(x－1)＝(x－1)[2(x－1)－1]＝(x－1)(2x－2－1)＝(x－1)(2x－3)－a⁴＋2a²b²－b⁴＝－(a⁴－2a²b²＋b⁴)＝－(a²－b²)²...
1) 2(x-1)^2-x+1=2(x^2-2x+1)-x+1=2x^2-5x+3=(2x-3)*(x-1)
2) -a^4+2a^2b^2-b^4=-(a^2-b^2)^2
3) (x+2y)^2-4(x-3y)^2=x^2+4xy+4y^2-4(x^2-6xy+9y^2)=-3x^2+28xy-32y^2=(-3x+4y)(x-8y)
4) 9(2x-y)^2-6(2x-y)+1=9(4x^2-4xy+y^2)-12x+6y+1=36x^2-36xy-12x+6y+1+9y^2
2(x-1)^2-x+1=(x-1)(2x-2-1)
-a^4+2a^2b^2-b^4=-(a^2-b^2)^2
(x+2y)^2-4(x-3y)^2=(x+2y+2x-6y)(x+2y-2x+6y)=(3x-4y)(8y-x)
9(2x-y)^2-6(2x-y)+1=[3(2x-y)-1]^2
(m^2-6m)^2+18(m^2-6m)+81=(m^2-6m+9)^2=(m-3)^4
-12a(a-b)^3+6a^2(b-a)^3=6a(b-a)^3(a+2)
When x is equal to something, the values of formula 4x + 3 and 3x + 4 are equal
In the range of real numbers, x equals 1
X=1
4x+3=3x+4
X=1
Factorization 8ab ^ 2-8a ^ 2b-2b ^ 3
solution
8ab²-8a²b-2b³
=2b(4ab-4a²-b²)
=-2b(4a²-4ab+b²)
=-2b(2a-b)²
8ab²-8a²b-2b³
=-2b(-4ab+4a²+b²)
=2b(2a-b)²
y²-6y-9x²+9
4x quartic-a & sup2; - 6a-9
36-b²-c²+2bc
ax²+bx²-cx²+a+b-c
A & sup2; - B & sup2; + 2 (AX - no) + X & sup2; - Y & sup2;
Y & sup2; - 6y-9x & sup2; + 9 = y & sup2; - 6y + 9-9x & sup2; = (Y-3) & sup2; - 9x & sup2; = (Y-3 + 3x) (y-3-3x) 4x quartic power - A & sup2; - 6a-9 = 4x quartic power - (A & sup2; + 6A + 9) = 4x quartic power - (a + 3) & sup2; = (2x & sup2; - A-3) (2x & sup2; + A + 3) 36-b & sup2; - C & sup2
1.（y-3）^2 - （3x）²=（y-3+3x）* （y-3 - 3x）
2. （2X ^2）^2 - (a+3)^2=(2X ^2+a+3) * (2X ^2-a-3)
3. 6^2 - (b+c)^2=(6+b+c) * (6-b+c)
4. Don't understand
The order is as follows:
y(y-6)+(3+3x)(3-3x)
(2x^2-a-3)(2x^2+a+3)
(6+b-c)(6-b+c)
(a+b-c)(x^2+1)
The last one should be a ^ 2-B ^ 2 + 2 (AX by) + x ^ 2-y ^ 2 = (a + x) ^ 2 - (B + y) ^ 2 = (a + x-b-y) (a + X + B + y)