\(2x^2+5x=3\)
\(2x^2+5x-3=0\)
\(2x^2-x+6x-3=0\)
\(\left(2x^2-x\right)+\left(6x-3\right)=0\)
\(x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(x+3\right)=0\)
\(2x-1=0\) hoặc \(x+3=0\)
*) \(2x-1=0\)
\(2x=1\)
\(x=\dfrac{1}{2}\)
*) \(x+3=0\)
\(x=0-3\)
\(x=-3\)
Vậy \(x=-3;x=\dfrac{1}{2}\)
\(2x^2+5x=3\)
\(\text{ }\Leftrightarrow2x^2+5x-3=0\)
\(\Leftrightarrow2x^2+6x-x-3=0\)
\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(2x^2+5x=3\)
\(\Leftrightarrow2x^2+5x-3=0\)
\(\Leftrightarrow2x^2+6x-x-3=0\)
\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
Vậy.........
2\(x^2\) + 5\(x\) = 3
2\(x^2\) + 5\(x\) - 3 = 0
2.(\(x^2\) + \(\dfrac{5}{2}\)\(x\) + \(\dfrac{25}{16}\)) - \(\dfrac{49}{8}\) = 0
2.(\(x\) + \(\dfrac{5}{4}\))2 = \(\dfrac{49}{8}\)
(\(x\) + \(\dfrac{5}{4}\))2 = \(\dfrac{49}{16}\)
\(\left[{}\begin{matrix}x+\dfrac{5}{4}=-\dfrac{7}{4}\\x+\dfrac{5}{4}=\dfrac{7}{4}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
