Đáp án:
a) \(\left[ \begin{array}{l}
x = - 2\\
x = 1\\
x = - 1\\
x = 0
\end{array} \right.\)
b) x=0
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{3}{{{x^2} + x + 1}}\\
A \in Z \to \dfrac{3}{{{x^2} + x + 1}} \in Z\\
\to {x^2} + x + 1 \in U\left( 3 \right);\left( {{x^2} + x + 1 > 0\forall x} \right)\\
\to \left[ \begin{array}{l}
{x^2} + x + 1 = 3\\
{x^2} + x + 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} + x - 2 = 0\\
{x^2} + x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left( {x + 2} \right)\left( {x - 1} \right) = 0\\
x\left( {x + 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\\
x = 1\\
x = - 1\\
x = 0
\end{array} \right.\\
b)DK:x \ge 0\\
B = \dfrac{{ - 6}}{{x + \sqrt x + 3}}\\
B \in Z \to \dfrac{6}{{x + \sqrt x + 3}} \in Z\\
\to x + \sqrt x + 3 \in U\left( 6 \right)\\
\to \left[ \begin{array}{l}
x + \sqrt x + 3 = 6\\
x + \sqrt x + 3 = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x + \sqrt x - 3 = 0\\
x + \sqrt x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = \dfrac{{ - 1 \pm \sqrt {13} }}{2}\left( l \right)\\
\sqrt x \left( {\sqrt x + 1} \right) = 0
\end{array} \right.\\
\to x = 0
\end{array}\)
