Đáp án:
$\begin{array}{l}
1.5.Dkxd:x \ne 0\\
a)A = \sqrt {\dfrac{{{{\left( {{x^2} - 3} \right)}^2} + 12{x^2}}}{{{x^2}}}} + \sqrt {{{\left( {x + 2} \right)}^2} - 8x} \\
= \sqrt {\dfrac{{{x^4} - 6{x^2} + 9 + 12{x^2}}}{{{x^2}}}} + \sqrt {{x^2} + 4x + 4 - 8x} \\
= \sqrt {\dfrac{{{x^4} + 6{x^2} + 9}}{{{x^2}}}} + \sqrt {{x^2} - 4x + 4} \\
= \sqrt {\dfrac{{{{\left( {{x^2} + 3} \right)}^2}}}{{{x^2}}}} + \sqrt {{{\left( {x - 2} \right)}^2}} \\
= \dfrac{{{x^2} + 3}}{{\left| x \right|}} + \left| {x - 2} \right|\\
+ Khi:x \ge 2\\
\Rightarrow A = \dfrac{{{x^2} + 3}}{x} + x - 2\\
+ Khi:0 < x < 2\\
\Rightarrow A = \dfrac{{{x^2} + 3}}{x} + 2 - x\\
+ Khi:x < 0\\
\Rightarrow A = \dfrac{{{x^2} + 3}}{{ - x}} + 2 - x\\
b)A = \dfrac{{{x^2} + 3}}{{\left| x \right|}} + \left| {x - 2} \right|\\
= \left| x \right| + \dfrac{3}{{\left| x \right|}} + \left| {x - 2} \right|\\
A \in Z\\
\Rightarrow \dfrac{3}{{\left| x \right|}} \in Z\\
\Rightarrow \left[ \begin{array}{l}
\left| x \right| = 1\\
\left| x \right| = 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = - 1\\
x = 3\\
x = - 3
\end{array} \right.\\
Vậy\,x \in \left\{ { - 3; - 1;1;3} \right\}
\end{array}$
