Question

Giải bất pt sau : -5 ( 2x-1 )( 4x+3 ) < 0

Answer 1

-5(2x-1)(4x+3)<0

=>(2x-1)(4x+3)>0

TH1: \(\left\{{}\begin{matrix}2x-1>0\\4x+3>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>\dfrac{1}{2}\\x>-\dfrac{3}{4}\end{matrix}\right.\)

=>\(x>\dfrac{1}{2}\)

TH2: \(\left\{{}\begin{matrix}2x-1< 0\\4x+3< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x< -\dfrac{3}{4}\end{matrix}\right.\)

=>\(x< -\dfrac{3}{4}\)

Answer 2

\(-5\left(2x-1\right)\left(4x+3\right)< 0\)

\(\Leftrightarrow\left(2x-1\right)\left(4x+3\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1>0\\4x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1< 0\\4x+3< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x>1\\4x>-3\end{matrix}\right.\\\left\{{}\begin{matrix}2x< 1\\4x< -3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{1}{2}\\x>-\dfrac{3}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x< -\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{2}\\x< -\dfrac{3}{4}\end{matrix}\right.\)

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