Đáp án:
\(\begin{array}{l}
a)x \in \emptyset \\
b)\left[ \begin{array}{l}
- \dfrac{{13}}{2} > x\\
x > - 2
\end{array} \right.\\
c)x \ge 2\\
d)\left[ \begin{array}{l}
x \ge 4\\
x \le - 1
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a){x^2} + 6x + 8 > {x^2} + 6x - 16 + 26\\
\to 8 > 10\left( {KTM} \right)\\
\to x \in \emptyset \\
b)DK:x \ne - 2\\
\dfrac{{ - 3}}{{x + 2}} < \dfrac{2}{3}\\
\to \dfrac{{ - 9 - 2\left( {x + 2} \right)}}{{3\left( {x + 2} \right)}} < 0\\
\to \dfrac{{ - 13 - 2x}}{{3\left( {x + 2} \right)}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
- 13 - 2x > 0\\
x + 2 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
- 13 - 2x < 0\\
x + 2 > 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
- \dfrac{{13}}{2} > x\\
x < - 2
\end{array} \right.\\
\left\{ \begin{array}{l}
- \dfrac{{13}}{2} < x\\
x > - 2
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
- \dfrac{{13}}{2} > x\\
x > - 2
\end{array} \right.\\
c){x^3} - 2{x^2} + x - 2 \ge 0\\
\to {x^2}\left( {x - 2} \right) + \left( {x - 2} \right) \ge 0\\
\to \left( {x - 2} \right)\left( {{x^2} + 2} \right) \ge 0\\
\to x - 2 \ge 0\left( {do:{x^2} + 2 > 0\forall x} \right)\\
\to x \ge 2\\
d)\left| {2x - 3} \right| \ge 5\\
\to \left[ \begin{array}{l}
2x - 3 \ge 5\\
2x - 3 \le - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x \ge 4\\
x \le - 1
\end{array} \right.
\end{array}\)
