Đáp án:
c) \(\left[ \begin{array}{l}
x = 2\\
x = - 4\\
x = 0\\
x = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \pm 1\\
A = \left[ {\dfrac{{x + 6 + 2\left( {x - 1} \right) + x\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right].\dfrac{{3\left( {x - 1} \right)}}{{x + 2}}\\
= \dfrac{{x + 6 + 2x - 2 + {x^2} + x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{3\left( {x - 1} \right)}}{{x + 2}}\\
= \dfrac{{{x^2} + 4x + 4}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{3\left( {x - 1} \right)}}{{x + 2}}\\
= \dfrac{{{{\left( {x + 2} \right)}^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{3\left( {x - 1} \right)}}{{x + 2}}\\
= \dfrac{{3\left( {x + 2} \right)}}{{x + 1}} = \dfrac{{3x + 6}}{{x + 1}}\\
b)\left| {x + 3} \right| = 4\\
\to \left[ \begin{array}{l}
x + 3 = 4\\
x + 3 = - 4
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = - 7
\end{array} \right.\\
Thay:x = - 7\\
\to A = \dfrac{{3\left( { - 7} \right) + 6}}{{ - 7 + 1}} = \dfrac{5}{2}\\
c)A = \dfrac{{3x + 6}}{{x + 1}} = \dfrac{{3\left( {x + 1} \right) + 3}}{{x + 1}}\\
= 3 + \dfrac{3}{{x + 1}}\\
A \in Z \Leftrightarrow \dfrac{3}{{x + 1}} \in Z\\
\Leftrightarrow x + 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 3\\
x + 1 = - 3\\
x + 1 = 1\\
x + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = - 4\\
x = 0\\
x = - 2
\end{array} \right.
\end{array}\)
