=>4x^2-5x+1=0
=>(x-1)(4x-1)=0
=>x=1 hoặc x=1/4
\(1,6x^2-2x+0,4=0\)
\(\Leftrightarrow\dfrac{8}{5}x^2-2x+\dfrac{2}{5}=0\)
\(\Leftrightarrow\dfrac{8}{5}x^2-\dfrac{8}{5}x-\dfrac{2}{5}x+\dfrac{2}{5}=0\)
\(\Leftrightarrow\dfrac{8}{5}x\left(x-1\right)-\dfrac{2}{5}\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{8}{5}x-\dfrac{2}{5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{8}{5}x-\dfrac{2}{5}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\dfrac{8}{5}x=\dfrac{2}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(S=\left\{1;\dfrac{1}{4}\right\}\)
\(1,6x^2-2x+0,4=0\)
\(\Leftrightarrow2,5\left(1,6x^2-2x+0,4\right)=0\)
\(\Leftrightarrow4x^2-5x+1=0\)
\(\Leftrightarrow4x^2-4x-x+1=0\)
\(\Leftrightarrow4x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{4}\end{matrix}\right.\)
