Question

Phân tích đa thức thành nhân tử:

x^10+x^5+1

Answer 1

x^10+x^5+1

=x¹⁰-x+x⁵-x²+x²+x+1

=x.(x⁹-1)+x².(x³-1)+(x²+x+1)

=x.(x³-1)(x³+1)+x²(x³-1)+(x²-x+1)

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

=(x²+x+1)[x.(x-1)(x³+1)+x²(x-1)+1]

=(x²+x+1)(x⁵+x²-x⁴-x+x³-x²+1)

=(x²+x+1)(x⁵-x⁴+x³-x+1)

Answer 2

 

x^10+x^5+1

=x¹⁰-x+x⁵-x²+x²-x+1

=x.(x⁹-1)+x².(x³-1)+(x²+x+1)

=x.(x³-1)(x³+1)+x²(x³-1)+(x²-x+1)

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

=(x²+x+1)[x.(x-1)(x³+1)+x²(x-1)+1]

=(x²+x+1)(x⁵+x²-x⁴-x+x³-x²+1)

=(x²+x+1)(x⁵-x⁴+x³-x+1)

 

Answer 3

sai rồi

Answer 4

ta có: 
A= x^10 + x^5 + 1 
A= (x^10 -x) + (x^5-x²) + (x²+x+1) 

A= x.(x³ -1).(x^6+x³+1) + x².(x³-1) + (x²+x+1) 
A= x.(x -1).(x²+x+1).(x^6+x³+1) + x².(x-1).(x²+x+1) + (x²+x+1) 

A= (x²+x+1).[x.(x-1).(x^6+x³+1) + x².(x²+x+1) +1] 

Answer 5

(x^3-1)(x^3+1)=x^6-1 mà

Answer 6

dễ mà bạn 

Answer 7

có cách tắt hơn nè L: ( cách cô giáo mk lm )

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

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

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

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

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