Đáp án:
2) \(\left[ \begin{array}{l}
x = 4\\
x = - 10\\
x = - 2\\
x = - 4
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)6\sqrt x \vdots 2\sqrt x - 3\\
\Leftrightarrow 3\left( {2\sqrt x - 3} \right) + 9 \vdots 2\sqrt x - 3\\
\Leftrightarrow 9 \vdots 2\sqrt x - 3\\
\Leftrightarrow 2\sqrt x - 3 \in U\left( 9 \right)\\
\to \left[ \begin{array}{l}
2\sqrt x - 3 = 9\\
2\sqrt x - 3 = - 9\left( l \right)\\
2\sqrt x - 3 = 3\\
2\sqrt x - 3 = - 3\\
2\sqrt x - 3 = 1\\
2\sqrt x - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x = 6\\
\sqrt x = 3\\
\sqrt x = 0\\
\sqrt x = 2\\
\sqrt x = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 36\\
x = 9\\
x = 0\\
x = 4\\
x = 1
\end{array} \right.\\
2)A = \dfrac{{1 - 2x}}{{x + 3}} = \dfrac{{ - 2\left( {x + 3} \right) + 7}}{{x + 3}}\\
= - 2 + \dfrac{7}{{x + 3}}\\
A \in Z \Leftrightarrow \dfrac{7}{{x + 3}} \in Z\\
\Leftrightarrow x + 3 \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
x + 3 = 7\\
x + 3 = - 7\\
x + 3 = 1\\
x + 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = - 10\\
x = - 2\\
x = - 4
\end{array} \right.
\end{array}\)
