Phân tích x^4+2016x^2+2015x+2016 thành nhân tử
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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)
\(x^4+2016x^2+2015x+2016\)
=\(x^4+x^3+x^2+2015x^2+2015x+2015+1-x^3\)
=\(x^2\left(x^2+x+1\right)+2015\left(x^2+x+1\right)+\left(1-x\right)\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^2+2015+1-x\right)\)
=\(\left(x^2+x+1\right)\left(x^2-x+2016\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2015x4 + 2016x2 + x + 2016
= (2015x4 + 2015x3 + 2015x2) + (- 2015x3 - 2015x2 - 2015x) + (2016x2 + 2016x + 2016)
= (x2 + x + 1)(2015x2 - 2015x + 2016)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta 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)
\(x^4+2016x^2+2015x+2016\)
\(=x^4-x+2016x^2+2016x+2016\)
\(=x\left(x^3-1\right)+2016\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2016\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=x^4-x+2016x^2+2016x+2016.\)
\(=x\left(x^3-1\right)+2016\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2016\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bạn nhóm ba số giữa vs nhau r lấy x^4+1 xong phân k ra hehe mk cx ko chắc
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(\Rightarrow\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}\)
\(\Rightarrow\begin{cases}a-1=0\\b-1=0\\c-1=0\end{cases}\)\(\Rightarrow\begin{cases}a=1\\b=1\\c=1\end{cases}\)
\(\Rightarrow a=b=c=1\Rightarrow H=1\cdot1\cdot1+1^{2014}+1^{2015}+1^{2016}=1+1+1+1=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 _@
![](https://rs.olm.vn/images/avt/0.png?1311)
x4+2015x2+2014x+2015
=x4-x+2015x2+2015x+2015
=x.(x3-1)+2015.(x2+x+1)
=x.(x-1)(x2+x+1)+2015.(x2+x+1)
=(x2+x+1)(x2-x+2015)
\(x^4+2015x^2+2014x+2015=\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(2015x^2+2015x+2015\right)\)
\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+2015\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2-x+2015\right)\)