Đáp án:
1. `(3 - x)(x + 1) < 0`
<=> \(\left[ \begin{array}{l}3 - x > 0 ; x + 1 < 0\\3 - x < 0 ; x + 1 > 0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x < 3 ; x < -1\\x > 3 ; x > -1\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x < -1\\x > 3\end{array} \right.\)
2 . `(3x - 1)(2x + 4) ≥ 0`
<=> \(\left[ \begin{array}{l}3x - 1 > 0 ; 2x + 4 > 0\\3x - 1 < 0 ; 2x + 4 < 0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x > 1/3 ; x > -2\\x < 1/3 ; x < -2\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x > 1/3\\x < -2\end{array} \right.\)
3. `(x + 5)/(x - 1) < 0`
<=> \(\left[ \begin{array}{l}x + 5 > 0 ; x - 1 < 0\\x + 5 < 0 ; x - 1 > 0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x > -5 ; x < 1\\x < -5 ; x > 1 < Loại >\end{array} \right.\)
`<=> -5 < x < 1`
Giải thích các bước giải:
