Ta có: \(\left(2x+1\right)\left(5x-1\right)=20x^2-16x-1\)
\(\Leftrightarrow10x^2-2x+5x-1-20x^2+16x+1=0\)
\(\Leftrightarrow-10x^2+19x=0\)
\(\Leftrightarrow x\left(-10x+19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-10x+19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-10x=-19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{19}{10}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{19}{10}\right\}\)
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