Đáp án:
$\begin{array}{l}
1)x = - \frac{1}{2}\left( {tm} \right)\\
\Rightarrow A = \frac{1}{{{{\left( { - \frac{1}{2}} \right)}^2}}} - \frac{1}{{{{\left( { - \frac{1}{2}} \right)}^3}}} = \frac{1}{{\frac{1}{4}}} + \frac{1}{{\frac{1}{8}}} = 4 + 8 = 12\\
2)B = \frac{1}{{{{\left( {x + 1} \right)}^3}}}\left( {\frac{1}{x} + 1} \right) + \frac{1}{{2{x^2} + 4x + 2}}.\left( {\frac{1}{{{x^2}}} + 1} \right)\\
= \frac{1}{{{{\left( {x + 1} \right)}^3}}}.\frac{{x + 1}}{x} + \frac{1}{{2{{\left( {x + 1} \right)}^2}}}.\frac{{{x^2} + 1}}{{{x^2}}}\\
= \frac{1}{{x{{\left( {x + 1} \right)}^2}}} + \frac{{{x^2} + 1}}{{2{x^2}{{\left( {x + 1} \right)}^2}}}\\
= \frac{{2x + {x^2} + 1}}{{2{x^2}{{\left( {x + 1} \right)}^2}}} = \frac{{{{\left( {x + 1} \right)}^2}}}{{2{x^2}{{\left( {x + 1} \right)}^2}}} = \frac{1}{{2{x^2}}}\\
\Rightarrow P = \frac{A}{B} = A.\frac{1}{B} = \frac{{x - 1}}{{{x^3}}}.2{x^2} = 2.\frac{{x - 1}}{x} = \frac{{2x - 2}}{x}\\
3)Dkxd:x \ne 0;x \ne - 1\\
P \le x\\
\Rightarrow \frac{{2x - 2}}{x} \le x\\
\Rightarrow \frac{{2x - 2}}{x} - x \le 0\\
\Rightarrow \frac{{2x - 2 - {x^2}}}{x} \le 0\\
\Rightarrow \frac{{{x^2} - 2x + 2}}{x} \ge 0\\
\Rightarrow \frac{{{{\left( {x - 1} \right)}^2} + 1}}{x} \ge 0\\
\Rightarrow x > 0\left( {do:{{\left( {x - 1} \right)}^2} + 1 > 0\forall x} \right)\\
Vậy\,x > 0
\end{array}$
