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x^10 + x^5 + 1 
= x^10 + x^9 - x^9 + x^8 - x^8 + x^7 - x^7 + x^6 - x^6 + x^5 + x^5 - x^5 + x^4 - x^4 + x^3 - x^3 + x^2 - x^2 + x - x + 1 
= (x^10 + x^9 + x^8) - (x^9 + x^8 + x^7) + (x^7 + x^6 + x^5) - (x^6 + x^5 + x^4) + (x^5 + x^4 + x^3) - (x^3 + x^2 + x) + (x^2 + x + 1) 
= x^8 (x^2 + x + 1) - x^7 (x^2 + x + 1) + x^5 (x^2 + x + 1) - x^4 (x^2 + x + 1) + x^3 (x^2 + x + 1) - x (x^2 + x + 1) + (x^2 + x + 1) 
= (x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1) 

\(a^{10}+a^5+1\)

\(=\left(a^{10}-a\right)+\left(a^5-a^2\right)+\left(a^2+a+1\right)\)

\(=a\left(a^3-1\right).\left(a^6+a^3+1\right)+a^2\left(a^3-1\right)+\left(a^2+a+1\right)\)

\(=a\left(a-1\right)\left(a^2+a+1\right)+\left(a^6+a^3+1\right)+a^2\left(a-1\right)\left(a^2+a+1\right)\)+  (a²+a+1) 

Đến đây rùi thì tự làm tiếp nha

\(a^{10}+a^5+1\)

\(=\left(a^{10}+a^9+a^8\right)-\left(a^9+a^8+a^7\right)+\left(a^7+a^6+a^5\right)-\left(a^6+a^5+a^4\right)+\left(a^5+a^4+a^3\right)-\left(a^3+a^2+a\right)+\left(a^2+a+1\right)\)\(=\left(a^2+a+1\right)\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

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

\(a^{10}+a^5+1\)

\(a^{10}+a^5+1=\left(a^2+a+1\right)\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

A = x⁸(x²-1)+1

A =(x²-1)(x⁸+1)

      \(x^{10}+x^8+1\)

\(=x^{10}-x+x^8-x^2+x^2+x+1\)

\(=x\left(x^9-1\right)+x^2\left(x^6-1\right)+x^2+x+1\)

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

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

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

\(=\left(x^2+x+1\right)\left(x-1\right)\left(x^7+x^5+x^4+x^2+x\right)+x^2+x+1\)

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

Chúc bạn học tốt.

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

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

\(a^5+a^3-a^2-1\)

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

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

trả lời thiếu
 

Ta có :

\(a^{10}+a^5+1\)

\(=a^{10}-a+a^5-a^2+\left(a^2+a+1\right)\)

\(=a\left(a^9-1\right)+a^2\left(a^3-1\right)+\left(a^2+a+1\right)\)

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

\(=\left(a^7+a+a^4\right)\left(a^3-1\right)+a^2\left(a^3-1\right)+\left(a^2+a+1\right)\)

\(=\left(a^3-1\right)\left(a^7+a^4+a^3+a^2+a+1\right)+\left(a^2+a+1\right)\)

\(=\left(a^2+a+1\right)\left(a-1\right)\left(a^7+a^4+a^3+a^2+a+1\right)+\left(a^2+a+1\right)\)

\(=\left(a^2+a+1\right)\left[\left(a-1\right)\left(a^7+a^4+a^3+a^2+a+1\right)+1\right]\)

\(=\left(a^2+a+1\right)\left(a^8-a^7+a^5+1\right)\)

 a¹⁰ + a⁵ + 1

= a¹⁰ - a⁹ + a⁷ - a⁶ + a⁵ - a³ + a² + a⁹ - a⁸ + a⁶ - a⁵ + a⁴ - a³ + a + a⁸ - a⁷ + a⁵ - a⁴ + a² - a + 1 

nhóm 7 hạng tử ta đc : 

= a²(a⁸ - a⁷ + a⁵ - a⁴ + a³ - a + 1) + a(a⁸ - a⁷ + a⁵ - a⁴ + a³ - a + 1) + (a⁸ - a⁷ + a⁵ - a⁴ + a³ - a + 1)

= (a² + a + 1)(a⁸ - a⁷ + a⁵ - a⁴ + a³ - a + 1)

                   = x.(x³ - 1).(x⁶ + x³ + 1) + x².(x³ - 1) + (x² + x + 1)

                   = (x² + x + 1). [x.(x -1).(x⁶ + x³ + 1) + x² + 1 ]