\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

\(x^8+x+1\)

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

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

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

\(=\left(x^6-x^5\right)\left(x^2+x+1\right)+\left(x^3-x^2\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

\(x^8+x+1\)

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

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

Đề đúng không thế. Nếu đúng thì bài này phức tạp lắm

\(x^8+3x^3+1\)

\(=x^8-x^4+4x^4+4\)

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

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

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

sai rồi bạn

x² + xy + 5x + 5y = ( x² + xy ) + ( 5x + 5y ) = x( x + y ) + 5( x + y ) = ( x + y )( x + 5 )

x² - y² + 3x - 3y = ( x² - y² ) + ( 3x - 3y ) = ( x - y )( x + y ) + 3( x - y ) = ( x - y )( x + y + 3 )

x² + xy + 5x + 5y 

= (x²+ xy) + ( 5x+5y)

= x(x+y) + 5(x+y)

= (x+y)(x+5)

x² - y² + 3x - 3y

= (x² - y²) + ( 3x -3y)

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

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

chúc bạn học tốt ^^

a) \(x^2+xy+5x+5y=\left(x^2+xy\right)+\left(5x+5y\right)\)

\(=x\left(x+y\right)+5\left(x+y\right)=\left(x+y\right)\left(x+5\right)\)

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

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

=x¹¹-x²+x²+x+1

=x²(x⁹-1)+(x²+x+1)

=x²[(x³)³-1³)+(x²+x+1)

=x²(x³-1)(x⁶+x³+1)+(x²+x+1)

=x²(x⁶+x³+1)(x-1)(x²+x+1)+(x²+x+1)

Đặt nhân tử chung là x²+x+1 rồi phá hết ngoặc là xong

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

\(x-\sqrt{x}-2\\ =x+\sqrt{x}-2\sqrt{x}-2\\ =\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}+1\right)\\ =\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)

\(x^3+27x+\left(x+3\right)\left(x-9\right)\)
⇒\(x^3+27x+x^2-6x-27\)
⇒\(x^3+x^2+21x-27\)
Chịu

Sửa đề: \(x^3+27+\left(x+3\right)\left(x-9\right)\)

\(=\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)\)

\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)

\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)

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

cá này là bình phương thếu.k thể phân tích thành nhân tử dc nữa

\(x\sqrt{x}-1=\sqrt{x^2}.x-1=\sqrt{x}^3-1=\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)