Đáp án:
$\begin{array}{l}
a)\frac{{x - 15}}{{2000}} + \frac{{x - 14}}{{2001}} + \frac{{x - 13}}{{2003}} = \frac{{x - 12}}{{2003}} + 2\\
\Rightarrow \frac{x}{{2000}} - \frac{{15}}{{2000}} + \frac{x}{{2001}} - \frac{{14}}{{2001}} + \frac{x}{{2003}} - \frac{{13}}{{2003}} = \frac{x}{{2003}} - \frac{{12}}{{2003}} + 2\\
\Rightarrow x.\left( {\frac{1}{{2000}} + \frac{1}{{2001}}} \right) = \frac{{15}}{{2000}} + \frac{{14}}{{2001}} + \frac{{13}}{{2003}} - \frac{{12}}{{2003}} + 2\\
\Rightarrow x = 2015,5\\
b)\left\{ \begin{array}{l}
{x^2} - 6x + 11 = {\left( {x - 3} \right)^2} + 2 \ge 2\\
{y^2} + 2y + 4 = {\left( {y + 1} \right)^2} + 3 \ge 3\\
2 + 4z - {z^2} = - {\left( {z - 2} \right)^2} + 6 \le 6
\end{array} \right.\\
\Rightarrow \left( {{x^2} - 6x + 11} \right)\left( {{y^2} + 2y + 4} \right) \ge 6\\
\Rightarrow \left( {{x^2} - 6x + 11} \right)\left( {{y^2} + 2y + 4} \right) = 2 + 4z - {z^2}\\
\Rightarrow \left\{ \begin{array}{l}
x = 3\\
y = - 1\\
z = 2
\end{array} \right.
\end{array}$
