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^7 + x^5 + 1
=x^7-x^6+x^5-x^3+x^2+x^6-x^5+x^4-x^2+x+x^5-x^4+x^3-x+1
=(x^2+x+1)(x^5-x^4+x^3-x+1)
+x^4 - 6x^3 + 12x^2 - 14x + 3
=x^4-2x^3+3x^2-4x^3-6x^2-12x+x^2-2x+3
=(x^2-4x+1)(x^2-2x+3)

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

Ta có :

x⁷ + x⁵ + 1 

= x⁷ + x⁶ - x⁶ + 2x⁵ - x⁵ + x⁴ - x⁴ + x³ - x³ + x² - x² +1

= x² . ( x⁵ - x⁴ + x³ - x + 1 ) + x . ( x⁵ - x⁴ + x³ - x + 1 ) + ( x⁵ - x⁴ + x³ - x + 1 )

= ( x² + x + 1 )( x⁵ - x⁴ + x³ - x + 1 )

Câu 4:

D=(x+1)(x+3)(x+5)(x+7)+15

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+105+15\)

\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+120\)

\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

\(=\left(x+2\right)\left(x+6\right)x^2+8x+10\)

Câu 2:

b: \(4x^2-12x+9\)

\(=\left(2x\right)^2-2\cdot2x\cdot3+3^2\)

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

Câu 1:

a: \(4x^2-9y^2=\left(2x\right)^2-\left(3y\right)^2=\left(2x-3y\right)\left(2x+3y\right)\)

b: \(\left(3x+y\right)^3=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot y+3\cdot3x\cdot y^2+y^3\)

\(=27x^3+27x^2y+9xy^2+y^3\)

Sửa đề: \(\frac13x^3+\frac72x^2+2x+1\)

\(\frac13x^3+\frac72x^2+2x+1\)

\(=\frac16\cdot21x^3+\frac16\cdot21x^2+\frac16\cdot12x+\frac16\cdot6\)

\(=\frac16\left(2x^3+21x^2+12x+6\right)\)

Biểu thức này không phân tích thành nhân tử.