\(2x^2-5x+3=0\)
\(\Rightarrow x^2-\dfrac{5}{2}x+\dfrac{3}{2}=0\)
\(\Rightarrow x^2-\dfrac{5}{4}x2+\dfrac{25}{16}-\dfrac{1}{16}=0\)
\(\Rightarrow\left(x-\dfrac{5}{4}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{4}=\dfrac{1}{4}\\x-\dfrac{5}{4}=\dfrac{-1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=1\end{matrix}\right.\)
Vậy \(x=\dfrac{3}{2}\) hoặc \(x=1\)
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