Đáp án:

\(A > 0\forall x \notin \left\{ { - \dfrac{3}{2};0; \pm 3} \right\}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)A = \dfrac{{3x}}{{x - 3}} - \dfrac{{{x^2} + 3x}}{{2x + 3}}.\left[ {\dfrac{{\left( {3x + 9} \right)\left( {x + 3} \right) - 3{x^2}}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}} \right]\text{ Đk: }x \notin \left\{ { - \dfrac{3}{2};0; \pm 3} \right\}\\
 = \dfrac{{3x}}{{x - 3}} - \dfrac{{{x^2} + 3x}}{{2x + 3}}.\dfrac{{3{x^2} + 18x + 27 - 3{x^2}}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{3x}}{{x - 3}} - \dfrac{{x\left( {x + 3} \right)}}{{2x + 3}}.\dfrac{{9\left( {2x + 3} \right)}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{3x}}{{x - 3}} - \dfrac{9}{{x - 3}} = \dfrac{{3\left( {x - 3} \right)}}{{x - 3}} = 3 > 0\forall x \notin \left\{ { - \dfrac{3}{2};0; \pm 3} \right\}\\
\end{array}\)

\(\begin{array}{l}
B = \dfrac{{1 - {x^2}}}{{2x}}.\dfrac{{{x^2} - x - 3}}{{x + 3}} + \dfrac{{3{x^2} - 14x + 3}}{{2x\left( {x + 3} \right)}}\text{ Đk: }x \notin \left\{ { -3;0} \right\}\\
 = \dfrac{{\left( {{x^2} - x - 3} \right)\left( {1 - {x^2}} \right) + 3{x^2} - 14x + 3}}{{2x\left( {x + 3} \right)}}\\
 = \dfrac{{ - {x^4} + {x^3} + 3{x^2} + {x^2} - x - 3 + 3{x^2} - 14x + 3}}{{2x\left( {x + 3} \right)}}\\
 = \dfrac{{ - {x^4} + {x^3} + 7{x^2} - 15x}}{{2x\left( {x + 3} \right)}}\\
 = \dfrac{{x\left( { - {x^3} + {x^2} + 7x - 15} \right)}}{{2x\left( {x + 3} \right)}}\\
 = \dfrac{{x\left( { - {x^3} - 3{x^2} + 4{x^2} + 12x - 5x - 15} \right)}}{{2x\left( {x + 3} \right)}}\\
 = \dfrac{{ - {x^2}\left( {x + 3} \right) + 4x\left( {x + 3} \right) - 5\left( {x + 3} \right)}}{{2\left( {x + 3} \right)}}\\
 = \dfrac{{\left( {x + 3} \right)\left( { - {x^2} + 4x - 5} \right)}}{{2\left( {x + 3} \right)}} = \dfrac{{ - {x^2} + 4x - 5}}{2}\\
 = \dfrac{{ - {x^2} + 4x - 5}}{2}\\
Do: - {x^2} + 4x - 5 =-(x-2)^2-1< 0\forall x\\
 \to \dfrac{{ - {x^2} + 4x - 5}}{2} < 0\\
 \to B < 0\forall x \notin \left\{ { - 3;0} \right\}
\end{array}\)