a) \(5x^2-4x+1=0\)
Ta có: \(5x^2-4x+1=5\left(x^2-\dfrac{4}{5}x+\dfrac{1}{5}\right)\)
= \(5\left(x^2-2x.\dfrac{2}{5}+\dfrac{4}{25}+\dfrac{1}{25}\right)\)
= \(5\left[\left(x-\dfrac{2}{5}\right)^2+\dfrac{1}{25}\right]\)
= \(5\left(x-\dfrac{2}{5}\right)^2+\dfrac{1}{5}>0\forall x\)
Do đó phương trình trên vô nghiệm.
b) \(x^2-x-6=0\)
\(\Leftrightarrow\) \(x^2-3x+2x-6=0\)
\(\Leftrightarrow\) \(x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\) \(\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy x = 3 hoặc x = -2
c) \(2x^2+x-1=0\)
\(\Leftrightarrow\) \(2x^2+2x-x-1=0\)
\(\Leftrightarrow\) \(2x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\) \(\left(x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy x = -1 hoặc x = \(\dfrac{1}{2}\)
