Tìm x ∈ ℤ, biết:

a) (x + 2)(x – 4) ≥ 0;

b) \(\frac{{2x + 3}}{{x + 4}} > 1\).

a) (x + 2)(x – 4) ≥ 0

\( \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x + 2 \ge 0\\x - 4 \le 0\end{array} \right.\\\left\{ \begin{array}{l}x + 2 \le 0\\x - 4 \ge 0\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge - 2\\x \le 4\end{array} \right.\\\left\{ \begin{array}{l}x \le - 2\\x \ge 4\end{array} \right.\end{array} \right.\)

⇔ −2 ≤ x ≤ 4

Vì x ∈ ℤ nên x ∈ {−2; −1; 0; 1; 2; 3; 4}.

b) \(\frac{{2x + 3}}{{x + 4}} < 1\)

\( \Leftrightarrow \frac{{2x + 3}}{{x + 4}} - 1 < 0\)

\( \Leftrightarrow \frac{{2x + 3 - x - 4}}{{x + 4}} < 0\)

\( \Leftrightarrow \frac{{x - 1}}{{x + 4}} < 0\)

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

⇔ −4 < x < 1

Vì x ∈ ℤ nên x ∈ {−3; −2; −1; 0}