Đáp án:
$\begin{array}{l}
P = \left( {\frac{{5x + 2}}{{{x^2} - 10}} + \frac{{5x - 2}}{{{x^2} + 10}}} \right).\frac{{{x^2} - 100}}{{{x^2} + 4}}\\
= \frac{{\left( {5x + 2} \right)\left( {{x^2} + 10} \right) + \left( {5x - 2} \right)\left( {{x^2} - 10} \right)}}{{\left( {{x^2} - 10} \right)\left( {{x^2} + 10} \right)}}.\frac{{{x^2} - 100}}{{{x^2} + 4}}\\
= \frac{{5{x^3} + 50x + 2{x^2} + 20 + 5{x^3} - 50x - 2{x^2} + 20}}{{{x^2} + 4}}\\
= \frac{{10{x^3} + 40}}{{{x^2} + 4}}\\
2)\\
\frac{3}{{2x - 16}} + \frac{{3x - 20}}{{x - 8}} + \frac{1}{8} = \frac{{13x - 102}}{{3x - 24}}\left( {dkxd:x \ne 8} \right)\\
\Rightarrow \frac{3}{{2\left( {x - 8} \right)}} + \frac{{3x - 20}}{{x - 8}} + \frac{1}{8} = \frac{{13x - 102}}{{3\left( {x - 8} \right)}}\\
\Rightarrow \frac{{3.4 + 8.\left( {3x - 20} \right) + x - 8}}{{8\left( {x - 8} \right)}} = \frac{{13x - 102}}{{3\left( {x - 8} \right)}}\\
\Rightarrow \frac{{25x - 156}}{8} = \frac{{13x - 102}}{3}\\
\Rightarrow 75x - 468 = 104x - 816\\
\Rightarrow x = 12\left( {tm} \right)
\end{array}$
Vậy x=12
