Đáp án:
\(\begin{array}{l}
a.\left\{ \begin{array}{l}
x \ne 2\\
x \ne - 2
\end{array} \right.\\
b.A = \frac{{x + 2}}{{x - 2}}\\
c.\left[ \begin{array}{l}
x = 3\\
x = 1\\
x = 4\\
x = 0\\
x = 6\\
x = - 2
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
a. ĐKXĐ: \(\left\{ \begin{array}{l}
x - 2 \ne 0\\
x + 2 \ne 0
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x \ne 2\\
x \ne - 2
\end{array} \right.\)
\(\begin{array}{l}
b.A = \left( {\frac{1}{{x - 2}} - \frac{1}{{x + 2}}} \right).\frac{{{x^2} + 4x + 4}}{4}\\
= \left[ {\frac{{x + 2}}{{(x - 2)(x + 2)}} - \frac{{x - 2}}{{(x - 2)(x + 2)}}} \right].\frac{{{{(x + 2)}^2}}}{4}\\
= \frac{{x + 2 - x + 2}}{{(x - 2)(x + 2)}}.\frac{{{{(x + 2)}^2}}}{4}\\
= \frac{{x + 2}}{{x - 2}}\\
c.A = \frac{{x + 2}}{{x - 2}} = \frac{{x - 2 + 4}}{{x - 2}} = 1 + \frac{4}{{x - 2}}
\end{array}\)
Để A đạt giá trị nguyên
<-> 4 chia hết cho x-2
\( \leftrightarrow \left[ \begin{array}{l}
x - 2 = 1\\
x - 2 = - 1\\
x - 2 = 2\\
x - 2 = - 2\\
x - 2 = 4\\
x - 2 = - 4
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 1\\
x = 4\\
x = 0\\
x = 6\\
x = - 2
\end{array} \right.\)
