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

ngu dốt

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

mik chịu

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)

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)

sai rồi

x¹⁰ + x⁵ + 1 = (x¹⁰ - x) + (x⁵ - x²) + (x² + x + 1) = x.[(x³)³ - 1] + x².(x³ - 1) + (x² + x + 1)

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

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

mình làm tương tự mấy ban kia nha

k tui nha

thannks

\(x+7\sqrt{x}+10=\left(\sqrt{x}+2\right)\left(\sqrt{x}+5\right)\)

= x^10 - x + x^5 - x^2 + x^2 + x + 1 

= x ( x^9 - 1 ) + x^2 (x^3 - 1 ) + x^2 + x + 1 

= x [ ( x^3 - 1) ( x^6 + x^3 + 1 )] + x^2 ( x - 1 )(x^2 + x + 1 ) + x^2 + x + 1 

= x  ( x - 1 )(x^2 + x + 1 )(x^6 + x^3 + 1) + x^2 (x-1 )(x^2 + x+  1 ) + x^2 + x + 1 

= (x^2 + x + 1 )[ x(x-1)(x^6 + x^3 + 1 ) + x^2 + 1 ) 

Nhân ra giúp mình nha 

vào câu hỏi  liên quan

\(x^{10}+x^2+1\)

\(=x^{10}-x^8+x^4+x^8-x^6+x^2+x^6-x^4+1\)

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

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

\(=\left(x^6-x^4+1\right)\left[x^4+2x^2+1-x^2\right]\)

\(=\left(x^6-x^4+1\right)\left[\left(x^2+1\right)^2-x^2\right]\)

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

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