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

x⁴ + x³ + x +1

= x³(x+1) + (x+1)

=(x+1)(x³ +1)

= (x+1)(x+1)(x² - x-1)

= (x+1)²(x² -x+1) 

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

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

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

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

bạn ơi  bạn viết sai đề có đề nào viết  thế này đâu?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

Uchiha Itachi

Ta có x 2 – 6x + 8

= x 2 – 4x – 2x + 8

= x(x – 4) – 2(x – 4)

= (x – 4)(x – 2)

Đáp án cần chọn là: A

\(16-8x+x^2\)
\(=4^2-2.4.x+x^2\)
\(=\left(4-x\right)^2\)

Ta có

x 3 y 3   +   6 x 2 y 2   +   12 x y   +   8 =   ( x y ) 3   +   3 ( x y ) 2 . 2   +   3 x y . 2 2   +   2 3   =   ( x y   +   2 ) 3

Đáp án cần chọn là: A

x³ - 3x² - 4x + 12

= x²(x-3) - 4(x-3)

= (x-3)(x² -4)

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

\(x^3-3x^2-4x+12\)

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

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

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

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