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 $x = \frac{3}{2}$.