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

\(A=\left(x-1\right)\left(x-2\right)\left(x+7\right)\left(x+8\right)+8\)

\(A=\left[\left(x-1\right)\left(x+7\right)\right]\left[\left(x-2\right)\left(x+8\right)\right]+8\)

\(A=\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8\)

Đặt \(q=x^2+6x-7\)ta có :

\(A=q\left(q-9\right)+8\)

\(A=q^2-9q+8\)

\(A=q^2-q-8q+8\)

\(A=q\left(q-1\right)-8\left(q-1\right)\)

\(A=\left(q-1\right)\left(q-8\right)\)

Thay \(q=x^2+6x-7\)vào A ta được :

\(A=\left(x^2+6x-7-1\right)\left(x^2+6x-7-8\right)\)

\(A=\left(x^2+6x-8\right)\left(x^2+6x-15\right)\)

bạn xem lại xem thử có sai đề bài ko

đề sai nha bạn

mình sửa đề cho:

\(A=\left(x+1\right)\left(x+2\right)\left(x+7\right)\left(x+8\right)+8\)

\(A=\left(x+1\right)\left(x+8\right)\left(x+2\right)\left(x+7\right)+8\)

\(A=\left(x^2+9x+8\right)\left(x^2+9x+14\right)+8\)

Đặt \(x^2+9x+8=a\)

\(\Rightarrow A=a\left(a+6\right)+8=a^2+6a+8=\left(a+2\right)\left(a+4\right)\)

\(\Rightarrow A=\left(x^2+9x+8+2\right)\left(x^2+9x+8+4\right)=\left(x^2+9x+10\right)\left(x^2+9x+12\right)\)

Mình nghĩ là bạn sai đề rồi.Mình có cách làm khác:

(x+1)(x+2)(x+7)(x+8)+8

=((x+1)(x+8))((x+2)(x+7))+8

=(x^2+9x+8)(x^2+9x+14)+8

=(x^2+9x+11-3)(x^2+9x+11+3)+8

=(x^2+9x+11)-9+8=(x^2+9x+11-1)(x^2+9x+11+1)

=(x^2+9x+10)(x^2+9x+12).

      \(x^8+x^7+1\)

\(=x^8-x^2+x^7-x+x^2+x+1\)

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

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

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

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

\(=\left(x^5+x^4+x^2+x\right)\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)

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

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

Chúc bạn học tốt.

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

a) x^7+x^2 +1 =x^7 - x^4+x^4 +x^2+1

                       = (x^7 - x^4) +[ (x^2)^2 +x^2 +1]

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

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

                       = (x-1)[ x^4( x^2+x+1)+(x+1)]

                       = (x-1)(x^6 +x^5+x^4+x+1)

b) x^8 +x+1 = x^8 -x^2+x^2 +x+1

                    = (x^8-x^2) +(x^2 +x+1)

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

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

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

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

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

Ta có :

\(x^8+x^7+1\)

\(=\left(x^8+x^7+x^6\right)-x^6+1\)

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

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

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

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

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

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

  x^9 + x^8 + x^7 - x^3 + 1 

= x^7 ( x^2 + x + 1 ) - ( x^3 - 1 )

= x^7 ( x^2 + x + 1 ) - ( x - 1 )(x^2 + x +   1 )

= ( x^7 - x + 1 )(x^2 + x + 1 )