Đáp án:
$\begin{array}{l}
B2)d)\\
\left( {x + 1} \right)\left( {x - 2} \right) < 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 < 0\\
x - 2 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 > 0\\
x - 2 > 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x < - 1\\
x < 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x > - 1\\
x > 2
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x < - 1\\
x > 2
\end{array} \right.\\
Vậy\,x < - 1,x > 2\\
B4)\\
A = 2,7 + \left| {3,4 - x} \right|\\
Do:\left| {3,4 - x} \right| \ge 0\\
\Leftrightarrow A = 2,7 + \left| {3,4 - x} \right| \ge 2,7\\
\Leftrightarrow GTNN:A = 2,7\,khi:x = 3,4
\end{array}$
