Đáp án:
d) \( - 2 < x < 3\)
Giải thích các bước giải:
\(\begin{array}{l}
B4:\\
a)3x + 2 \in B\left( {x + 1} \right)\\
\to 3x + 2 \vdots x + 1\\
\to 3\left( {x + 1} \right) - 1 \vdots x + 1\\
\to 1 \vdots x + 1\\
\to x + 1 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 1\\
x + 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = - 2
\end{array} \right.\\
b)4x - 2 \in B\left( {3x + 1} \right)\\
\to 4x - 2 \vdots 3x + 1\\
\to 12x - 6 \vdots 3x + 1\\
\to 4\left( {3x + 1} \right) - 10 \vdots 3x + 1\\
\to 10 \vdots 3x + 1\\
\to \left[ \begin{array}{l}
3x + 1 = 10\\
3x + 1 = - 10\\
3x + 1 = 5\\
3x + 1 = - 5\\
3x + 1 = 2\\
3x + 1 = - 2\\
3x + 1 = 1\\
3x + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\\
x = - \dfrac{{11}}{3}\left( l \right)\\
x = \dfrac{4}{3}\left( l \right)\\
x = - 2\\
x = \dfrac{1}{3}\left( l \right)\\
x = - 1\\
x = 0\\
x = - \dfrac{2}{3}\left( l \right)
\end{array} \right.\\
c)3x + 1 \in B\left( {2x + 2} \right)\\
\to 3x + 1 \vdots 2x + 2\\
\to 6x + 2 \vdots 2x + 2\\
\to 3\left( {2x + 2} \right) - 4 \vdots 2x + 2\\
\to 4 \vdots 2x + 2\\
\to 2x + 2 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
2x + 2 = 4\\
2x + 2 = - 4\\
2x + 2 = 2\\
2x + 2 = - 2\\
2x + 2 = 1\\
2x + 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\\
x = - 3\\
x = 0\\
x = - 2\\
x = - \dfrac{1}{2}\left( l \right)\\
x = - \dfrac{3}{2}\left( l \right)
\end{array} \right.\\
d)\left( {x + 2} \right)\left( {x - 3} \right) < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 > 0\\
x - 3 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 < 0\\
x - 3 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > - 2\\
x < 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x < - 2\\
x > 3
\end{array} \right.\left( l \right)
\end{array} \right.\\
\to - 2 < x < 3
\end{array}\)
