Đáp án:
$\begin{array}{l}
a)A = \dfrac{{11}}{{x + 3}}\\
\Rightarrow GTLN:A = 11\\
\Rightarrow x + 3 = 1\\
\Rightarrow x = - 2\\
b)B = \dfrac{{{5^2}}}{{{{10}^2}}} + \dfrac{{{5^2}}}{{{{11}^2}}} + \dfrac{{{5^2}}}{{{{12}^2}}} + ... + \dfrac{{{5^2}}}{{{{99}^2}}}\\
= {5^2}.\left( {\dfrac{1}{{{{10}^2}}} + \dfrac{1}{{{{11}^2}}} + ... + \dfrac{1}{{{{99}^2}}}} \right)\\
= {5^2}.C\\
Do:\dfrac{1}{{{{10}^2}}} > \dfrac{1}{{10.11}}\\
\dfrac{1}{{{{11}^2}}} > \dfrac{1}{{11.12}};\\
...\dfrac{1}{{{{99}^2}}} > \dfrac{1}{{99.100}}\\
\Rightarrow \dfrac{1}{{{{10}^2}}} + \dfrac{1}{{{{11}^2}}} + \dfrac{1}{{{{12}^2}}} + .. + \dfrac{1}{{{{99}^2}}}\\
> \dfrac{1}{{10.11}} + \dfrac{1}{{11.12}} + \dfrac{1}{{12.13}} + .. + \dfrac{1}{{99.100}}\\
\Rightarrow C > \dfrac{1}{{10}} - \dfrac{1}{{11}} + \dfrac{1}{{11}} - \dfrac{1}{{12}} + ... + \dfrac{1}{{99}} - \dfrac{1}{{100}}\\
\Rightarrow C > \dfrac{1}{{10}} - \dfrac{1}{{100}} = \dfrac{9}{{100}}\\
\Rightarrow {5^2}.C > {5^2}.\dfrac{9}{{100}}\\
\Rightarrow B > \dfrac{9}{4}
\end{array}$
