Đáp án:
$\begin{array}{l}
a)M = \dfrac{1}{{4{x^2} - 4x + 7}}\\
4{x^2} - 4x + 7\\
= 4{x^2} - 4x + 1 + 6\\
= {\left( {2x - 1} \right)^2} + 6 \ge 6\\
\Rightarrow \dfrac{1}{{4{x^2} - 4x + 7}} \le \dfrac{1}{6}\\
\Rightarrow M \le \dfrac{1}{6}\\
\Rightarrow GTNN:M = \dfrac{1}{6}\,khi:x = \dfrac{1}{2}\\
b)E = \dfrac{{2016}}{{4{x^2} + 16x + 19}}\\
4{x^2} + 16x + 19\\
= 4.\left( {{x^2} + 4x + 4} \right) + 3\\
= 4.{\left( {x + 2} \right)^2} + 3 \ge 3\\
\Rightarrow \dfrac{1}{{4{x^2} + 16x + 19}} \le \dfrac{1}{3}\\
\Rightarrow \dfrac{{2016}}{{4{x^2} + 16x + 19}} \le \dfrac{{2016}}{3} = 672\\
\Rightarrow E \le 672\\
\Rightarrow GTLN:E = 672\,khi:x = - 2\\
c)M = \dfrac{{{a^2} - 2a + 2016}}{{{a^2}}}\\
= 1 - \dfrac{2}{a} + \dfrac{{2016}}{{{a^2}}}\\
= 2016.\left( {\dfrac{1}{{{a^2}}} - 2.\dfrac{1}{a}.\dfrac{1}{{2016}} + \dfrac{1}{{{{2016}^2}}}} \right) - \dfrac{1}{{2016}} + 1\\
= 2016.{\left( {\dfrac{1}{a} - 1} \right)^2} + \dfrac{{2015}}{{2016}} \ge \dfrac{{2015}}{{2016}}\\
\Rightarrow M \ge \dfrac{{2015}}{{2016}}\\
\Rightarrow GTNN:M = \dfrac{{2015}}{{2016}}\,khi:\dfrac{1}{a} = 1 \Rightarrow a = 1\\
d)F = \dfrac{{{x^2}}}{{{x^2} - 2x + 2002}}\\
\Rightarrow F.{x^2} - 2F.x + 2002.F = {x^2}\\
\Rightarrow \left( {F - 1} \right).{x^2} - 2F.x + 2002.F = 0\\
DK:\Delta ' \ge 0\\
\Rightarrow {F^2} - 2002.F.\left( {F - 1} \right) \ge 0\\
\Rightarrow {F^2} - 2002{F^2} + 2002F \ge 0\\
\Rightarrow 2001{F^2} - 2002F \le 0\\
\Rightarrow F\left( {2001F - 2002} \right) \le 0\\
\Rightarrow 0 \le F \le \dfrac{{2002}}{{2001}}\\
\Rightarrow GTLN:F = \dfrac{{2002}}{{2001}}\,khi:x = 2002
\end{array}$
