Bài 1 :

\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)

Bài 2 :

 \(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)

\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)

=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)

Tick đúng nha 

X^2-6+8

a: =4(x-2)(x+1)+4(x-2)^2+(x+1)^2

=(2x-4)^2+2*(2x-4)(x+1)+(x+1)^2

=(2x-4+x+1)^2=(3x-3)^2=9(x-1)^2

b: =x^7(x^2-1)-x^5(x+1)+x^3(x+1)+(x^2-1)

=(x+1)[x^7(x-1)-x^5+x^3+x-1]

=(x+1)[x^7(x-1)-x^3(x-1)(x+1)+(x-1)]

=(x+1)(x-1)(x^7-x^4-x^3+1)

=(x+1)(x-1)(x^3-1)(x^4-1)

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

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

56.B

57.B

58.B

56B

57B

58B

M = x⁹ - x⁷ + x⁶ - x⁵ - x⁴ + x³ - x² + 1

= ( x⁹ - x⁷ ) + ( x⁶ - x⁴ ) - ( x⁵ - x³ ) - ( x² - 1 )

= x⁷( x² - 1 ) + x⁴( x² - 1 ) - x³( x² - 1 ) - ( x² - 1 )

= ( x² - 1 )( x⁷ + x⁴ - x³ - 1 )

= ( x - 1 )( x + 1 )[ x⁴( x³ + 1 ) - ( x³ + 1 ) ]

= ( x - 1 )( x + 1 )( x³ + 1 )( x⁴ - 1 )

= ( x - 1 )( x + 1 )( x + 1 )( x² - x + 1 )( x² - 1 )( x² + 1 )

= ( x + 1 )²( x - 1 )( x² - x + 1 )( x - 1 )( x + 1 )( x² + 1 )

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

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

\(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)

\(=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)\)

\(=\left(x+1\right)\left(-2x^2+3x+7\right)\)

\(=x\left(x-6\right)+2\left(x-6\right)=\left(x-6\right)\left(x+2\right)\)

ta có 

[(x+6)(x+7)+(4+x)(x+7)]+(x+4)(x+5)

=[(x+7)(x+6+x+4)]+(x+4)(X+5)

=(x+7)(2x+10)+(x+4)(x+5)

=(x+7)2(x+5)+(x+4)(x+5)

=(x+5)(2x+14+x+4)

=(x+5)(3x+18)

(x+6).(x+7)+(x+4).(x+7)+(x+4).(x+5)

= x^2+13x+42+x^2+11x+28+x^2+9x+20

= 3x^2+33x+90

= 3.(x^2+11x+30)

=3.[(x^2+5x)+(6x+30)]

= 3.(x+5).(x+6)

Tk mk nha

(x + 6)(x + 7) + (x + 4)(x + 7) + (x + 4)(x + 5)

= (x + 7)(x + 6 + x + 4) + (x + 4)(x + 5)

= (x + 7)(2x + 10) + (x + 4)(x + 5)

= 2(x + 7)(x + 5) + (x + 4)(x + 5)

= 2(x + 5)(x + 7 + x + 4)

= 2(x + 5)(2x + 11)

Lời giải:
\(x^3(x^2-7)^2-36x=x[x^2(x^2-7)^2-36]\\ =x[(x^3-7x)^2-6^2]=x(x^3-7x-6)(x^3-7x+6)\\ =x[x^2(x-3)+3x(x-3)+2(x-3)][x^2(x-2)+2x(x-2)-3(x-2)]\\ =x(x-3)(x^2+3x+2)(x-2)(x^2+2x-3)\\ =x(x-3)(x+1)(x+2)(x-2)(x-1)(x+3)\)