Đáp án:
$\begin{array}{l}
Đkxđ:\left\{ \begin{array}{l}
x \ne - 3\\
{x^2} + x - 6 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne - 3\\
\left( {x - 2} \right)\left( {x + 3} \right) \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne - 3\\
x \ne 2
\end{array} \right.\\
a)A = \frac{{x + 2}}{{x + 3}} - \frac{5}{{{x^2} + x - 6}}\\
= \frac{{x + 2}}{{x + 3}} - \frac{5}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\
= \frac{{\left( {x + 2} \right)\left( {x - 2} \right) - 5}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\
= \frac{{{x^2} - 4 - 5}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\
= \frac{{{x^2} - 9}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\
= \frac{{\left( {x + 3} \right)\left( {x - 3} \right)}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\
= \frac{{x - 3}}{{x - 2}}\\
b)x \ne 2;x \ne - 3\\
A > 0\\
\Rightarrow \frac{{x - 3}}{{x - 2}} > 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 3 > 0\\
x - 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3 < 0\\
x - 2 < 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 3\\
x > 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 3\\
x < 2
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x > 3\\
x < 2
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x > 3\\
x < 2;x \ne - 3
\end{array} \right.\\
c)x \ne - 3;x \ne 2\\
A = \frac{{x - 3}}{{x - 2}} = \frac{{x - 2 - 1}}{{x - 2}} = 1 - \frac{1}{{x - 2}}
\end{array}$
Để A nguyên dương thì 1/(x-2) phải nguyên và $\frac{1}{{x - 2}} < 1$
$ \Rightarrow \left[ \begin{array}{l}
x - 2 = 1\\
x - 2 = - 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
x = 1\left( {tm} \right)
\end{array} \right.$
Thử lại + x=3 thì $A = \frac{{x - 3}}{{x - 2}} = 0$ ko nguyên dương
=> x=3 loại
+x=1 thì $A = \frac{{x - 3}}{{x - 2}} = 2$ nguyên dương
Vậy x=1
