Đáp án:
$\begin{array}{l}
Dkxd:x \ne 1;x \ne \dfrac{3}{2}\\
P = \left( {\dfrac{{2x}}{{2{x^2} - 5x + 3}} - \dfrac{5}{{2x - 3}}} \right):\left( {3 + \dfrac{2}{{1 - x}}} \right)\\
= \left( {\dfrac{{2x}}{{\left( {2x - 3} \right)\left( {x - 1} \right)}} - \dfrac{5}{{2x - 3}}} \right)\\
:\dfrac{{3\left( {1 - x} \right) + 2}}{{1 - x}}\\
= \dfrac{{2x - 5\left( {x - 1} \right)}}{{\left( {2x - 3} \right)\left( {x - 1} \right)}}.\dfrac{{1 - x}}{{3 - 3x + 2}}\\
= \dfrac{{ - 3x + 5}}{{2x - 3}}.\dfrac{{ - 1}}{{ - 3x + 5}}\\
= \dfrac{{ - 1}}{{2x - 3}}\\
= \dfrac{1}{{3 - 2x}}\\
a)P > 0\\
\Leftrightarrow \dfrac{1}{{3 - 2x}} > 0\\
\Leftrightarrow 3 - 2x > 0\\
\Leftrightarrow x < \dfrac{3}{2}\\
Vậy\,x < \dfrac{3}{2};x \ne 1\\
b)P = \dfrac{1}{{6 - {x^2}}}\\
\Leftrightarrow \dfrac{1}{{3 - 2x}} = \dfrac{1}{{6 - {x^2}}}\\
\Leftrightarrow 3 - 2x = 6 - {x^2}\\
\Leftrightarrow {x^2} - 2x - 3 = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow x = 3;x = - 1\left( {tm} \right)\\
Vậy\,x = 3;x = - 1
\end{array}$
