Đáp án:
$\begin{array}{l}
1)Do:{x^2} \ge 0\\
\Rightarrow - {x^2} \le 0\\
\Rightarrow - {x^2} + 4 \le 4\\
\Rightarrow \dfrac{1}{{ - {x^2} + 4}} \ge \dfrac{1}{4}\\
\Rightarrow C \ge \dfrac{1}{4}\\
\Rightarrow \min C = \dfrac{1}{4}\\
D = - \dfrac{1}{{{x^2} - 4}}\\
{x^2} - 4 \ge - 4\\
\Rightarrow \dfrac{1}{{{x^2} - 4}} \le - \dfrac{1}{4}\\
\Rightarrow - \dfrac{1}{{{x^2} - 4}} \ge \dfrac{1}{4}\\
\Rightarrow D \ge \dfrac{1}{4}\\
\Rightarrow MinD = \dfrac{1}{4}\\
2) - {x^2} + 2x - 3\\
= - {x^2} + 2x - 1 - 2\\
= - \left( {{x^2} - 2x + 1} \right) - 2\\
= - {\left( {x - 1} \right)^2} - 2 \le - 2\\
\Rightarrow \dfrac{1}{{ - {x^2} + 2x - 3}} \ge - \dfrac{1}{2}\\
\Rightarrow \dfrac{4}{{ - {x^2} + 2x - 3}} \ge - 2\\
\Rightarrow \min E = - 2\\
F = \dfrac{{ - 4}}{{{x^2} - 2x + 3}}\\
{x^2} - 2x + 3 = {\left( {x - 1} \right)^2} + 2 \ge 2\\
\Rightarrow \dfrac{1}{{{x^2} - 2x + 3}} \le \dfrac{1}{2}\\
\Rightarrow \dfrac{{ - 4}}{{{x^2} - 2x + 3}} \ge - 4\\
\Rightarrow F \ge - 4\\
\Rightarrow \min \,F = - 4\\
3)H = \dfrac{3}{{ - {x^2} - 4x + 1}}\\
- {x^2} - 4x + 1\\
= - {x^2} - 4x - 4 + 5\\
= - \left( {{x^2} + 4x + 4} \right) + 5\\
= - {\left( {x + 2} \right)^2} + 5 \le 5\\
\Rightarrow \dfrac{3}{{ - {x^2} - 4x + 1}} \ge \dfrac{3}{5}\\
\Rightarrow \min H = \dfrac{3}{5}
\end{array}$
