Phân tích đa thức sau thành nhân tử: x^10 + x^2015 +1
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^{10}+x^8+1\)
\(=x^{10}-x+x^8-x^2+x^2+x+1\)
\(=x\left(x^9-1\right)+x^2\left(x^6-1\right)+x^2+x+1\)
\(=x\left(x^3-1\right)\left(x^6+x^3+1\right)+x^2\left(x^3+1\right)\left(x^3-1\right)+x^2+x+1\)
\(=\left(x^7+x^4+x\right)\left(x^3-1\right)+\left(x^5+x^2\right)\left(x^3-1\right)+x^2+x+1\)
\(=\left(x^3-1\right)\left(x^7+x^5+x^4+x^2+x\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x-1\right)\left(x^7+x^5+x^4+x^2+x\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^8-x^7+x^6-x^4+x^3-x+1\right)\)
Chúc bạn học tốt.
* là nhân à ???????????
\(x^2-x-2015\times2016=\left(x^2-2016x\right)+\left(2015x-2015\times2016\right)=x.\left(x-2016\right)+2015\left(x-2016\right)=\left(x-2016\right)\left(x+2015\right)\)k nha
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(x^{10}+x^2+1\)
\(=x^{10}-x^8+x^4+x^8-x^6+x^2+x^6-x^4+1\)
\(=x^4\left(x^6-x^4+1\right)+x^2\left(x^6-x^4+1\right)+\left(x^6-x^4+1\right)\)
\(=\left(x^6-x^4+1\right)\left(x^4+x^2+1\right)\)
\(=\left(x^6-x^4+1\right)\left[x^4+2x^2+1-x^2\right]\)
\(=\left(x^6-x^4+1\right)\left[\left(x^2+1\right)^2-x^2\right]\)
\(=\left(x^6-x^4+1\right)\left(x^2+1+x\right)\left(x^2+1-x\right)\)
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
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