Giải thích các bước giải:

ĐKXĐ: $x\ne\pm3, x\ne 2$

Ta có:

$\dfrac{x+3}{x-2}-\dfrac{x-3}{x+3}=\dfrac{48x^3}{x^2-9}$

$\to \left(x+3\right)^2\left(x-3\right)-\left(x-2\right)\left(x-3\right)^2=48x^3\left(x-2\right)$

$\to 11x^2-30x-9=48x^4-96x^3$

$\to 48x^4-96x^3+9+30x-11x^2=0$

$\to \left(4x-3\right)\left(12x^3-15x^2-14x-3\right)=0$

$\to 4x-3=0\quad \mathrm{hoặc}\quad \:12x^3-15x^2-14x-3=0$

$\to x=\dfrac{3}{4},\:x\approx \:1.92393\dots $