Ta có: \(5x^2-6x+1=0\)
\(\Rightarrow5x^2-5x-x+1=0\)
\(\Rightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{5};x=1\) là nghiệm của đa thức \(5x^2-6x+1\)
Ta có: \(5x^2-6x+1\)
\(=5\left(x^2-\dfrac{6}{5}x+\dfrac{1}{5}\right)\)
\(=5\left(x^2-\dfrac{3}{5}x-\dfrac{3}{5}x+\dfrac{1}{5}\right)\)
\(=5\left(x^2-\dfrac{3}{5}x-\dfrac{3}{5}x+\dfrac{9}{25}-\dfrac{4}{25}\right)\)
\(=5\left[x\left(x-\dfrac{3}{5}\right)-\dfrac{3}{5}\left(x-\dfrac{3}{5}\right)-\dfrac{4}{25}\right]\)
\(=5\left[\left(x-\dfrac{3}{5}\right)^2-\dfrac{4}{25}\right]\)
Khi đó: \(5\left[\left(x-\dfrac{3}{5}\right)^2-\dfrac{4}{25}\right]=0\)
\(\Rightarrow\left(x-\dfrac{3}{5}\right)^2-\dfrac{4}{25}=0\)
\(\Rightarrow\left(x-\dfrac{3}{5}\right)^2=\dfrac{4}{25}\)
\(\Rightarrow\left(x-\dfrac{3}{5}\right)^2=\left(\dfrac{2}{5}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{5}=\dfrac{2}{5}\\x-\dfrac{3}{5}=\dfrac{-2}{5}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(x=1;x=\dfrac{1}{5}\) là \(n_o\) của đa thức.