Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ne 0\\
x + 2 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne 0\\
x \ne - 2
\end{array} \right.\\
b)Q = \frac{{{{\left( {x + 2} \right)}^2}}}{x}.\left( {1 - \frac{{{x^2}}}{{x + 2}}} \right) - \frac{{{x^2} + 6x + 4}}{x}\\
= \frac{{{{\left( {x + 2} \right)}^2}}}{x}.\frac{{x + 2 - {x^2}}}{{x + 2}} - \frac{{{x^2} + 6x + 4}}{x}\\
= \frac{{\left( {x + 2} \right)\left( { - {x^2} + x + 2} \right)}}{x} - \frac{{{x^2} + 6x + 4}}{x}\\
= \frac{{ - {x^3} - {x^2} + 4x + 4 - {x^2} - 6x - 4}}{x}\\
= \frac{{ - {x^3} - 2{x^2} - 2x}}{x}\\
= - {x^2} - 2x - 2\\
c)Q = - {x^2} - 2x - 2\\
= - \left( {{x^2} + 2x + 2} \right)\\
= - \left[ {{{\left( {x + 1} \right)}^2} + 1} \right]\\
Do:{\left( {x + 1} \right)^2} + 1 \ge 1 > 0\forall x \ne 0;x \ne - 2\\
\Rightarrow - \left[ {{{\left( {x + 1} \right)}^2} + 1} \right] < 0\\
\Rightarrow Q < 0\forall x \ne 0;x \ne - 2\\
d)Q = - \left[ {{{\left( {x + 1} \right)}^2} + 1} \right] \le 1\forall x \ne 0;x \ne - 2\\
\Rightarrow GTLN:Q = 1 \Leftrightarrow x = - 1
\end{array}$
