2015x⁴ + 2016x² + x + 2016

= (2015x⁴ + 2015x³ + 2015x²) + (- 2015x³ - 2015x² - 2015x) + (2016x² + 2016x + 2016)

= (x² + x + 1)(2015x² - 2015x + 2016)

Vào câu trả lời tương tự đi có đáp án đó

bạn có: x^4 + 2016x^2 + 2015x + 2016
= x^4 + x^3 + x^2 - x^3 - x^2 - x + 2016x^2 + 2016x + 2016
= x^2(x^2 + x + 1) - x(x^2 + x + 1) + 2016(x^2 + x + 1)
= (x^2 + x + 1)(x^2 - x + 2016)

  \(x^4+2016x^2+2015x+2016\)

=\(x^4+x^3+x^2+2015x^2+2015x+2015+1-x^3\)

=\(x^2\left(x^2+x+1\right)+2015\left(x^2+x+1\right)+\left(1-x\right)\left(x^2+x+1\right)\)

=\(\left(x^2+x+1\right)\left(x^2+2015+1-x\right)\)

=\(\left(x^2+x+1\right)\left(x^2-x+2016\right)\)

bạn nhóm ba số giữa vs nhau r lấy x^4+1 xong phân k ra hehe mk cx ko chắc

Ta có : x^4+2017x^2+2016x+2017

=x^4+x^3-x^3+x^2-x^2+2017x^2+2017x-x+2017

=x^4+x^3+x^2-x^3-x^2-x+2017x^2+2017x+2017

=x^2(x^2+x+1)-x(x^2+x+1)+2017(x^2+x+1)

=(x^2+x+1)(x^2-x+2017)

Nhớ k mk nha

Ta có : x^4+2017x^2+2016x+2017
=x^4+x^3-x^3+x^2-x^2+2017x^2+2017x-x+2017
=x^4+x^3+x^2-x^3-x^2-x+2017x^2+2017x+2017
=x^2(x^2+x+1)-x(x^2+x+1)+2017(x^2+x+1)
=(x^2+x+1)(x^2-x+2017)

chúc cậu hok tốt _@

\(x^4+2015x^2+2014x+2015.\)

=\(\left(x^4-x\right)+2015x^2+2015x+2015\)

=\(x\left(x^3-1\right)+2015\left(x^2+x+1\right)\)

=\(x\left(x-1\right)\left(x^2+x+1\right)+2015\left(x^2+x+1\right)\)

= \(\left(x^2+x+1\right)\left(x^2-x-2015\right)\)

k cho mik

a)\(x^2+7x+6\)

\(=x^2+6x+x+6\)

\(=x\left(x+6\right)+\left(x+6\right)\)

\(=\left(x+1\right)\left(x+6\right)\)

b)\(x^4+2016x^2+2015x+2016\)

\(=x^4+2016x^2+\left(2016x-x\right)+2016\)

\(=\left(x^4-x\right)+\left(2016x^2+2016x+2016\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2016\right)\)

Bài 3:

Từ \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)

\(\Rightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)

\(\Rightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)

\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\) (1)

Ta thấy:\(\begin{cases}\left(a-1\right)^2\ge0\\\left(b-1\right)^2\ge0\\\left(c-1\right)^2\ge0\end{cases}\)

\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\) (2)

Từ (1) và (2) \(\Rightarrow\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}\)

\(\Rightarrow\begin{cases}a-1=0\\b-1=0\\c-1=0\end{cases}\)\(\Rightarrow\begin{cases}a=1\\b=1\\c=1\end{cases}\)

\(\Rightarrow a=b=c=1\Rightarrow H=1\cdot1\cdot1+1^{2014}+1^{2015}+1^{2016}=1+1+1+1=4\)

Sửa đề: 2016x^2

x^4+2016x^2+2015x+2016

=x^4+x^3+x^2-x^3-x^2-x+2016x^2+2016x+2016

=(x^2+x+1)(x^2-x+2016)