a: \(2y\left(x+2\right)-3x-6\)

\(=2y\left(x+2\right)-3\left(x+2\right)\)

\(=\left(x+2\right)\left(2y-3\right)\)

b: \(3\left(x+4\right)-x^2-4x\)

\(=3\left(x+4\right)-\left(x^2+4x\right)\)

\(=3\left(x+4\right)-x\left(x+4\right)\)

\(=\left(x+4\right)\left(3-x\right)\)

c: \(2\left(x+5\right)-x^2-4x\)

\(=2x+10-x^2-4x\)

\(=-x^2-2x+10\)

\(=-x^2-2x-1+11\)

\(=11-\left(x^2+2x+1\right)\)

\(=11-\left(x+1\right)^2\)

\(=\left(\sqrt{11}-x-1\right)\left(\sqrt{11}+x+1\right)\)

d: \(x^2+6x-3x-18\)

\(=\left(x^2+6x\right)-\left(3x+18\right)\)

\(=x\left(x+6\right)-3\left(x+6\right)\)

\(=\left(x+6\right)\left(x-3\right)\)

\(x^2+4x+4=\left(x+2\right)^2 \)

\(4x^2-4x+1=\left(2x-1\right)^2\)

\(c\left(x+1\right)-y\left(x+1\right)=\left(x+1\right)\left(c-y\right)\)

\(x^3-3x^2+3x-1+27y^3=\left(x-1\right)^3+27y^3=\left(x-1+3y\right)\left(x^2-2x+1-3xy+3y+9y^2\right)\)

\(1,a^2-b^2-12a+12b=\left(a-b\right)\left(a+b\right)-12\left(a-b\right)=\left(a-b\right)\left(a+b-12\right)\\ 2,4x^2-4x+1-25y^2=\left(2x-1\right)^2-\left(5y\right)^2=\left(2x-5y-1\right)\left(2x+5y-1\right)\\ c,x^2-3x-10=\left(x^2-5x\right)+\left(2x-10\right)=x\left(x-5\right)+2\left(x-5\right)=\left(x-5\right)\left(x+2\right)\)

Phân tích đa thức thành nhân tử(tách hạng tử)
1)x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)=(x-1)(x+3)
2)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
3)x^2+7x+12=(x+3)(x+4)
4)x^2-x-12=(x-4)(x+3)
5)3x^2+3x-36=3[(x-3)(x+4)]
6)5x^2-5x-10=5[(x-2)(x+1) ]       
7)3x^2-7x-6=(x-3)(3x+2)
8)4x^2+4x-3=4x^2+6x-2x-3=(2x-1)(2x+3)
9)8x^2-2x-3=8x^2+4x-6x-3=(4x-3)(2x+1)
 

1: \(x^2+2x-3=\left(x+3\right)\left(x-1\right)\)

2: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

3: \(x^2+7x^2+12x=4x\left(2x+3\right)\)

4: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

5: \(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)

6: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)

x^2 - 7xy + 10y^2 
= (x^2 - 2xy) - (5xy - 10y^2) 
= x(x - 2y) - 5y( x - 2y) 
= (x - 5y)(x - 2y) 

mik mới học lớp 7 thôi!

Ta có : x³ - 3x² + 1 - 3x

= x³ - 2x² - x - (x² - 2x - 1) 

= x(x² - 2x - 1) - (x² - 2x - 1) 

= (x² - 2x - 1) (x - 1) 

\(H=\left(x^2-x+1\right)\left(x^2+3x+1\right)+4x^2\)

Đặt \(x^2+1=t\), ta được:

\(H=\left(t-x\right)\left(t+3x\right)+4x^2\)

\(H=t^2+2xt+x^2\)

\(H=\left(t+x\right)^2\)

\(H=\left(x^2+x+1\right)^2\)

Vậy.......

\(1,x^3+2x^2y+xy^2-4x\)

\(x\left(x^2+2xy+y^2-4\right)\)

\(x\left[\left(x+y\right)^2-2^2\right]\)

\(x\left(x+y+2\right)\left(x+y-2\right)\)

\(2,5x-5y-x^2+2xy-y^2\)

\(5\left(x-y\right)-\left(x^2-2xy+y^2\right)\)

\(5\left(x-y\right)-\left(x-y\right)^2\)

\(\left(x-y\right)\left(5-x+y\right)\)

\(3,x^4-3x^2\)

\(x^2\left(x^2-3\right)\)

a)x²-5x-14=(x²-7x)+(2x-14)=x(x-7)+2(x-7)=(x-7)(x+2)

b)4x²-3x-1=(4x²-4x)+(x-1)=4x(x-1)+(x-1)=(x-1)(4x+1)

c)x²-7xy+12y²=x²-7xy+12,25y²-0,25y²=(x-3,5y)²-0,25y²=(x-3,5y-0,5y)(x-3,5y+0,5y)=(x-4y)(x-2y)

d)x³-3x+2=(x³-x)-(2x-2)=(x-1)(x²+x)-2(x-1)=(x-1)(x²+x-2)=(x-1)(x²-2x+x-2)=(x-1)(x+1)(x-2)