\(5x^2+7x+2=0\)
\(\Rightarrow5x^2+5x+2x+2=0\)
\(\Rightarrow5x.\left(x+1\right)+2.\left(x+1\right)=0\)
\(\Rightarrow\left(5x+2\right).\left(x+1\right)=0\)
\(\Rightarrow5x+2=0\) hoặc \(x+1=0\)
\(\Rightarrow x=\dfrac{-2}{5}\) hoặc x=-1
\(5x^2+7x+2=0\)
\(\Leftrightarrow5x^2+5x+2x+2=0\)
\(\Leftrightarrow5x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(5x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{5}\\x=-1\end{matrix}\right.\)
Vậy \(x=\dfrac{-2}{5}\) hoặc x = -1
Ta có :
\(5x^2+7x+2=0\)
\(5x^2+7x=0-2\)
\(5x^2+7x=-2\)
\(5x.x+7x=-2\)
\(x\left(5x-7x\right)=-2\)
\(x.\left(-2\right)x=-2\)
\(x.x=1\)
\(\Leftrightarrow x=1\) hoặc \(x=-1\)
Vậy \(x\in\left\{1;-1\right\}\)
P/S : ko chắc...Chắc sai!
