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

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

k mk nha

bạn ơi bạn chưa bớt 2x^2 kìa

x⁵-x⁴-1=x⁵-x³-x²-x⁴+x²+x+x³-x-1

=x².(x³-x-1)-x.(x³-x-1)+(x³-x-1)

=(x³-x-1)(x²-x+1)

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

k mk nha

a; \(x^4+2x^3+10x^2-20x\)

\(=x\cdot x^3+x\cdot2x^2+x\cdot10x-x\cdot20\)

\(=x\left(x^3+2x^2+10x-20\right)\)

b: \(x^3-x^2y-xy^2+y^3\)

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

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

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

c: \(x^5+x^3-x^2-1\)

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

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

\(x^2\left(x-1\right)+16\left(1-x\right)\)

\(=x^2\left(x-1\right)-16\left(x-1\right)\)

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

\(=\left(x-1\right)\left(x-4\right)\left(x-4\right)\)

=x²(x-1)-16(x-1)

=(x-1) (x²-16)

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

\(3x\left(x-1\right)-4x\left(x-1\right)\)

\(3x\left(x-1\right)-4x\left(x-1\right)\)

\(-x\left(x-1\right)\)

\(-1x^2+x\)

\(-x^2+x\)

\(3x\left(x-1\right)-4x\left(x-1\right)\)

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

\(-1\left(x-1\right)x\)

\(3x\left(x-1\right)-4x\left(x-1\right)\)

\(=\left(3x-4x\right)\left(x-1\right)\)

\(=x\left(x-1\right)\)

 Dùng phương pháp: Đặt nhân tử chung

Còn không ghi là (x^2 + 3x + 1)^2