1) x³ + 2x² + x

= x(x² + 2x + 1)

= x(x + 1)²

2) 5x³ - 10x² + 5x

= 5x(x² - 2x + 1)

= 5x(x - 1)²

3) 8x²y - 8xy + 2x

= 2x(4xy - 4y + 1)

5) 2x² + 5x³ + x²y

= x²(2 + 5x + y)

6) 4x²y - 8xy² + 18x²y²

= 2xy(2x - 4y + 9xy) 

  Chúc Bạn Học Tốt !

a) 20x - 5y

= 5(4x - y)

b) 5x(x - 1)- 3x(x - 1)

= 2x(x - 1)

c) x(x + y) - 6x - 6y

= x(x + y) - (6x + 6y)

= x(x + y) - 6(x + y)

= (x + y)(x - 6)

d) 6x³ - 9x²

= 3x²(2x - 3)

e) 4x²y - 8xy² + 10x²y²

= 2xy(2x - 4y + 5xy)

g) 20x²y - 12x³

= 4x²(5y - 3x)

h) 8x⁴ + 12x²y⁴ - 16x³y⁴

= 4x²(2x² + 3y⁴ - 4xy⁴)

k) 4xy² + 8xyz

= 4xy(y + 2z)

giúp mình vs ạ c ơn nhiều

\(1,=8xy+14y^2-4xz-7yz\\ 2,=y\left(4x^2-12x+9\right)=y\left(2x-3\right)^2\\ 3,\Leftrightarrow\left(x+3\right)\left(x-2+x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

Câu 1: \(\left(2y-z\right)\left(4x+7y\right)=8xy-4xz+14y^2-7yz\)

câu 2: \(4x^2y-12xy+9y=y\left(4x^2-12x+9\right)\)

câu 3: \(\left(x-2\right)\left(x+3\right)+x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-2+x\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2x-2\right)=0\\ \Leftrightarrow2\left(x+3\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

1.

$4x^2y+5x^3-x^2y^2=x^2(4y+5x-y^2)$

2.

$5x(x-1)-3y(1-x)=5x(x-1)+3y(x-1)=(x-1)(5x+3y)$

3.

$4x^2-25=(2x)^2-5^2=(2x-5)(2x+5)$

4.

$6x-9-x^2=-(x^2-6x+9)=-(x-3)^2$

5.

$x^2+4y^2+4xy=x^2+2.x.2y+(2y)^2=(x+2y)^2$

6.

$\frac{1}{64}-27x^3=(\frac{1}{4})^3-(3x)^3$
$=(\frac{1}{4}-3x)(\frac{1}{16}+\frac{3x}{4}+9x^2)$
 

7.

$x^3-6x^2+12x-8=x^3-3.x^2.2+3.x.2^2-2^3$

$=(x-2)^3$
8.

$x^2-x-y^2-y=(x^2-y^2)-(x+y)=(x-y)(x+y)-(x+y)$

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

9.

$5x-5y+ax-ay=5(x-y)+a(x-y)$

$=(x-y)(5+a)$

a, 2xy^2 ( x^3 -3xy - 4 )

b, x^2 - 4x - 4x +16

= x(x-4) - 4(x-4)

= (x-4) (x-4)

sao có 2 câu v bạn :v

\(1,\\ 12x^6y^3:4x^3y=3x^3y^2\\ \left(x+1\right)\left(x^2-x+1\right)=x^3+1\\ 2x^2y\left(x^2+3xy\right)=3x^4y+6x^3y^2\\ 2,\\ a,=2xy\left(2x+3y-4\right)\\ b,=\left(x-3\right)\left(x+y\right)\\ c,=\left(x-2\right)\left(x+2\right)+y\left(x-2\right)=\left(x+y+2\right)\left(x-2\right)\\ d,=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\\ 3,\\ a,\Leftrightarrow x^2-x^2+2x=2\\ \Leftrightarrow2x=2\Leftrightarrow x=1\\ b,\Leftrightarrow\left(x-2\right)\left(x-2+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(=3xy\left(x+2y-6xy\right)\)

\(=3xy\left(x+2y-6xy\right)\)

a: \(3xy^2-3x^3-6xy+3x\)

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

\(=3x\left\lbrack\left(y-1\right)^2-x^2\right\rbrack\)

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

b: \(3x^2+11x+6\)

\(=3x^2+9x+2x+6\)

=3x(x+3)+2(x+3)

=(x+3)(3x+2)

c: \(-x^3-4xy^2+4x^2y+16x\)

\(=x\left(-x^2-4y^2+4xy+16\right)\)

\(=x\left\lbrack4^2-\left(x^2-4xy+4y^2\right)\right\rbrack\)

\(=x\cdot\left\lbrack4^2-\left(x-2y\right)^2\right\rbrack\)

=x(4-x+2y)(4+x-2y)

d: \(xz-x^2-yz+2xy-y^2\)

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

=z(x-y)-\(\left(x-y^{}\right)^2\)

=(x-y)(z-x+y)

e: \(4x^2-y^2-6x+3y\)

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

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

\(-8x^3+1=1^3-\left(2x\right)^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)