Lời giải

Đáp án đúng là: D

Ta có \[\frac{2}{{{\rm{x}} + 3}} + \frac{3}{{{{\rm{x}}^2} - 9}}\]\[ = \frac{2}{{{\rm{x}} + 3}} + \frac{3}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\]

\[ = \frac{{2\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \frac{3}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\]

\[ = \frac{{2\left( {x - 3} \right) + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \frac{{2x - 6 + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \frac{{2x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\]

Mà \[\frac{2}{{{\rm{x}} + 3}} + \frac{3}{{{{\rm{x}}^2} - 9}} = 0\] nên \[\frac{{2x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = 0\]

\[2x - 3 = 0\]

\[2x = 3\]

\[x = \frac{3}{2}\]

Vậy \[x = \frac{3}{2}\].