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

Biểu thức này không phân tích thành nhân tử.

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)

\(=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)\)

\(=\left(x+1\right)\left(-2x^2+3x+7\right)\)

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)

\(x^7+x^5+1\)

\(=\left(x^7-x\right)\left(x^5-x^2\right)+\left(x^2+x+1\right)\)

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

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

\(=\left(x^2+x+1\right)\left[\left(x^5-x^4+x^2-x\right)+\left(x^3-x^2\right)+1\right]\)

\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)

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 )