17)

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

18)

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

19)

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

20)

\(x^2+x-20\\ =x^2+5x-4x-20\\ =x\left(x+5\right)-4\left(x+5\right)\\ =\left(x+5\right)\left(x-4\right)\)

\(17,x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)

\(18,3\left(x+4\right)-x^2-4x=3\left(x+4\right)-x\left(x+4\right)=\left(x+4\right)\left(3-x\right)\)

\(19,x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)\)

\(20,x^2+x-20=x^2-4x+5x-20=x\left(x-4\right)+5\left(x-4\right)=\left(x-4\right)\left(x+5\right)\)

Ta nhắc lại: Phương trình bậc hai phân tích được thành nhân tử khi và chỉ khi nó tồn tại nghiệm.

Ta thấy: `x^2-4x+12=(x-2)^2+8>=8>0AAx` nên ta không thể phân tích nhân tử cho phương trình này.

x² - 4x - 12

= x² + 2x - 6x - 12

= (x² + 2x) - (6x + 12)

= x(x + 2) - 6(x + 2)

= (x + 2)(x - 6) 

`x^2-4x+12`

`=x^2+2x-6x+12`

`=x(x+2)-6(x+2)`

`=(x+2)(x-6)`

\(x^2-8x+12=\left(x^2-6x\right)-\left(2x-12\right)=x\left(x-6\right)-2\left(x-6\right)=\left(x-2\right)\left(x-6\right)\)

=(x-2)(x-6)

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

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

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

\(3x^2-6xy+9y^2-12\)

\(=3\cdot x^2-3\cdot2xy+3\cdot3y^2-3\cdot4\)

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

=3*x^2-3*2xy+3*3y^2-3*4

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

\(4x\left(x-5\right)^2-12\left(4-x\right)^2\)

\(=4\left[x\left(x-5\right)^2-3\left(4-x\right)^2\right]\)

\(=4\left[x\left(x^2-10x+25\right)-3\left(16-8x+x^2\right)\right]\)

\(=4\left(x^3-10x^2+25x-48+24x-3x^2\right)\)

\(=4\left(x^3-13x^2+49x-48\right)\)

#Ayumu

\(x^2+2xy+y^2-x-y-12\)

\(=\left(x+y\right)^2-\left(x+y\right)-12\)

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

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

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

=(x^2+x)^2+4(x^2+x)-12

=(x^2+x+6)(x^2+x-2)

=(x^2+x+6)(x+2)(x-1)

x² - x - 12

= x² + 3x - 4x - 12

= x(x + 3) - 4(x + 3)

= (x - 4)(x + 3)

Trả lời:

x² - x - 12 

= x² - 4x + 3x - 12

= x ( x - 4 ) + 3 ( x - 4 )

= ( x + 3 ) ( x - 4 )