Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x\# - 2\\
x\# 0
\end{array} \right.\\
b)B = \left( {1 - \dfrac{{{x^2}}}{{x + 2}}} \right).\dfrac{{{{\left( {x + 2} \right)}^2}}}{x} - \dfrac{{{x^2} + 6x + 4}}{x}\\
= \dfrac{{x + 2 - {x^2}}}{{x + 2}}.\dfrac{{{{\left( {x + 2} \right)}^2}}}{x} - \dfrac{{{x^2} + 6x + 4}}{x}\\
= \dfrac{{ - {x^2} + x + 2}}{x}.\left( {x + 2} \right) - \dfrac{{{x^2} + 6x + 4}}{x}\\
= \dfrac{{\left( { - {x^2} + x + 2} \right)\left( {x + 2} \right) - {x^2} - 6x - 4}}{x}\\
= \dfrac{{ - {x^3} - 2{x^2} + {x^2} + 2x + 2x + 4 - {x^2} - 6x - 4}}{x}\\
= \dfrac{{ - {x^3} - 2{x^2} - 2x}}{x}\\
= - {x^2} - 2x - 2\\
c)B = - {x^2} - 2x - 2\\
= - \left( {{x^2} + 2x + 1} \right) + 1 - 2\\
= - {\left( {x + 1} \right)^2} - 1 \le - 1\\
\Leftrightarrow B \le - 1\\
\Leftrightarrow GTLN:B = - 1\,khi:x = - 1\left( {tmdk} \right)
\end{array}$
