\(2016x^2-x-2017=0\\ \Leftrightarrow2016x^2+2016x-2017x-2017=0\\ \Leftrightarrow2016x\left(x+1\right)-2017\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(2016x-2017\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\2016x-2017=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{2017}{2016}\end{matrix}\right.\)

tk mk đi

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 _@

a: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

\(x^4+2017x^2+2016x+2017\)

\(=\left(x^4+x^2+1\right)+2016\left(x^2+x+1\right)\)

\(=\left(x^4+2x^2+1-x^2\right)+2016\left(x^2+x+1\right)\)

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

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

\(=\left(x^2+x+1\right)\left(x^2-x+2017\right)\)

\(x^4+2017x^2+2016x+2017\)

\(=\left(x^4-x\right)+\left(2007x^2+2007x+2007\right)\)

\(=x.\left(x^3-1\right)+2007.\left(x^2+x+1\right)\)

\(=x.\left(x-1\right)\left(x^2+x+1\right)+2007.\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2007\right)\)

\(a,x^4-4x^3+x^2-4x=0\)

\(\Rightarrow\left(x^4-4x^3\right)+\left(x^2-4x\right)=0\)

\(\Rightarrow x^3\left(x-4\right)+x\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(x^2+x\right)=0\)

\(\Rightarrow x\left(x-4\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-1\end{matrix}\right.\)

\(b,x^3-5x^2+4x-20=0\)

\(\Rightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)

\(\Rightarrow x^2\left(x-5\right)+4\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x^2+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\)

\(\Rightarrow x=5\)

a) \(x^4-4x^3+x^2-4x=0\)

\(\Leftrightarrow\left(x^4-4x^3\right)+\left(x^2-4x\right)=0\)

\(\Leftrightarrow x^3\left(x-4\right)+x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^3+x\right)=0\)

\(\Leftrightarrow x\left(x-4\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x^2+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x^2=-1\left(loai\right)\end{matrix}\right.\)

Vậy x=0; x=4

b) \(x^3-5x^2+4x-20=0\)

\(\Leftrightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)

\(\Leftrightarrow x^2\left(x-5\right)+4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=-4\left(loai\right)\end{matrix}\right.\)

Vậy x=5