Đáp án:
$\begin{array}{l}
2)\\
a)\left( {x + 2} \right)\left( {x - 5} \right) + x\left( {1 - x} \right) = 2\\
\Rightarrow {x^2} - 5x + 2x - 10 + x - {x^2} = 2\\
\Rightarrow - 2x = 10 + 2\\
\Rightarrow - 2x = 12\\
\Rightarrow x = - 6\\
b)A = 2009 - 4{x^2} - {y^2} + 4x - 6y\\
= - 4{x^2} + 4x - 1 - {y^2} - 6y - 9 + 2019\\
= - \left( {4{x^2} - 4x + 1} \right) - \left( {{y^2} + 6y + 9} \right) + 2019\\
= - {\left( {2x - 1} \right)^2} - {\left( {y + 3} \right)^2} + 2019 \le 2019\forall x,y\\
Dấu = xảy\,ra \Leftrightarrow \left\{ \begin{array}{l}
2x - 1 = 0\\
y + 3 = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = \frac{1}{2}\\
y = - 3
\end{array} \right.\\
4)\\
a)Dkxd:{x^2} + 2x \ne 0\\
\Rightarrow x\left( {x + 2} \right) \ne 0\\
\Rightarrow x \ne 0;x \ne - 2\\
C = 0\\
\Rightarrow \frac{{{x^2} - 4}}{{{x^2} + 2x}} = 0\\
\Rightarrow {x^2} - 4 = 0\\
\Rightarrow {x^2} = 4\\
\Rightarrow \left[ \begin{array}{l}
x = 2\left( {tm} \right)\\
x = - 2\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 2\\
b)D = \frac{{{x^2}}}{{x - 6}}\left( {\frac{{{x^2} + 36}}{x} - 12} \right) + 2028\\
= \frac{{{x^2}}}{{x - 6}}.\frac{{{x^2} - 12x + 36}}{x} + 2028\\
= x.\frac{{{{\left( {x - 6} \right)}^2}}}{{x - 6}} + 2028\\
= x\left( {x - 6} \right) + 2028\\
= {x^2} - 6x + 9 + 2019\\
= {\left( {x - 3} \right)^2} + 2019 \ge 2019\forall x\\
Dấu = xảy\,ra \Leftrightarrow x = 3\left( {tmdk} \right)
\end{array}$
