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