Đáp án:
a) \(A = \dfrac{{ - 2x - 2}}{{x - 2}}\)
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{a)A = \left( {\dfrac{x}{{{x^2} - 4}} + \dfrac{2}{{2 - x}} - \dfrac{1}{{x + 2}}} \right).\left( {x + 2} \right)}\\
{ = \dfrac{{x - 2\left( {x + 2} \right) - x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\left( {x + 2} \right)}\\
{ = \dfrac{{ - 2x - 4 + 2}}{{x - 2}}}\\
\begin{array}{l}
= \dfrac{{ - 2x - 2}}{{x - 2}}\\
Thay:x = - 2\left( l \right)\\
Thay:x = \dfrac{1}{2}\\
\to A = \dfrac{{ - 2.\dfrac{1}{2} - 2}}{{\dfrac{1}{2} - 2}} = 2
\end{array}\\
{b)A = - \dfrac{{2x + 2}}{{x - 2}} = - \dfrac{{2\left( {x - 2} \right) + 6}}{{x - 2}}}\\
{ = - 2 - \dfrac{6}{{x - 2}}}\\
{A \in Z \Leftrightarrow \dfrac{6}{{x - 2}} \in Z}\\
{ \Leftrightarrow x - 2 \in U\left( 6 \right)}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x - 2 = 6}\\
{x - 2 = {\rm{\;}} - 6}\\
{x - 2 = 3}\\
{x - 2 = {\rm{\;}} - 3}\\
{x - 2 = 2}\\
{x - 2 = {\rm{\;}} - 2}\\
{x - 2 = 1}\\
{x - 2 = {\rm{\;}} - 1}
\end{array}} \right. \to \left[ {\begin{array}{*{20}{l}}
{x = 8}\\
{x = - 4}\\
{x = 5}\\
{x = - 1}\\
{x = 4}\\
{x = 0}\\
{x = 3}\\
{x = 1}
\end{array}} \right.}
\end{array}\)
