Đặt \(x^2+3x+1=t\)

\(\Rightarrow\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6=t.\left(t+1\right)-6\)

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

\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)

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

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

\(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)

Đặt \(x^2+3x+1=a\)ta có :

\(a\left(a+1\right)-6\)

\(=a^2+a-6\)

\(=a^2+6a-a-6\)

\(=\left(a^2+6a\right)-\left(a+6\right)\)

\(=a\left(a+6\right)-\left(a+6\right)\)

\(=\left(a+6\right)\left(a-1\right)\)

Thay \(a=x^2+3x+1\)vào A ta có :

\(A=\left(x^2+3x+1+6\right)\left(x^2+3x+1-1\right)\)

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

\(3x\left(x-2\right)-x+2+5x\left(x-2\right)=\left(x-2\right)\left(8x-1\right)\)

\(3x\left(x-2\right)-x+2+5x\left(x-2\right)=3x\left(x-2\right)-\left(x-2\right)+5x\left(x-2\right)=\left(x-2\right)\left(3x=1+5x\right)=\left(x-2\right)\left(8x-1\right)\)

Ta có: \(x^2-3x-28\)

\(=x^2-7x+4x-28\)

\(=x\left(x-7\right)+4\left(x-7\right)\)

\(=\left(x-7\right)\left(x+4\right)\)

\(x^3-3x^2+6x-4\)

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

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

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

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

x^3 - 3x^2 + 6x - 4 

<=> x^3-3x^2+3x-1+3x-3

<=>(x-1)^3+3(x-1)

<=>(x-1)+((x-1)^2+3)

<=>(x-1)+(x^2-2x+4)

a: \(=3\left(x^2-y^2-x+y\right)\)

\(=3\left[\left(x-y\right)\left(x+y\right)-\left(x-y\right)\right]\)

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

b: =(x-4)(x+1)

c: =x(x-1)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

jup mình với mọi người

júp mink với ace ơi

( x + 2 ) ( x² - 2x ) - 3x - 6

= ( x + 2 ) ( x² - 2x ) - ( 3x + 6 )

= ( x + 2 ) ( x² - 2x ) - 3 ( x + 2 )

= ( x + 2 ) ( x² - 2x - 3 )

= ( x + 2 ) [ ( x² + x ) - ( 3x + 3 ) ]

= ( x + 2 ) [ x ( x + 1 ) - 3 ( x + 1 ) ]

= ( x + 2 ) ( x - 3 ) ( x + 1 ) 

( x + 2 )( x² - 2x ) - 3x - 6

= ( x + 2 )( x² - 2x ) - 3( x + 2 )

= ( x + 2 )( x² - 2x - 3 )

= ( x + 2 )[ ( x² - 2x + 1 ) - 4 ]

= ( x + 2 )[ ( x - 1 )² - 2² ]

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

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