Question

giải bpt \(2x^2-3x-2\le0\)

Answer 1

\(2x^2-3x-2\le0\Leftrightarrow2x^2-4x+x-2\le0\Leftrightarrow2x.\left(x-2\right)+\left(x-2\right)\le0\)

\(\Leftrightarrow\left(2x+1\right).\left(x-2\right)\le0\Leftrightarrow\hept{\begin{cases}2x+1\ge0\\x-2\ge0\end{cases}\text{hoặc }}\hept{\begin{cases}2x+1\le0\\x-2\le0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge-\frac{1}{2}\\x\ge2\end{cases}}\text{hoặc }\hept{\begin{cases}x\le-\frac{1}{2}\\x\le2\end{cases}\Leftrightarrow\orbr{\begin{cases}x\ge2\\x\le-\frac{1}{2}\end{cases}}}\)