Điều kiện: $x^2\neq 8, x^2\neq 9,x^2\neq 11,x^2\neq 12$$\frac{6}{x^2-9}-\frac{7}{x^2-8}+\frac{4}{x^2-11}-\frac{3}{x^2-12}=0$
$\Leftrightarrow \frac{15-x^2}{(x^2-9)(x^2-8)}+\frac{x^2-15}{(x^2-11)(x^2-12)}=0$
$\Leftrightarrow (15-x^2)[\frac{1}{(x^2-9)(x^2-8)}-\frac{1}{(x^2-11)(x^2-12)}]=0$
$\Leftrightarrow x^2=15$ hay $(x^2-9)(x^2-8)=(x^2-11)(x^2-12)$
$\Leftrightarrow x^2=15$ hay $x^2=10$
Vậy $x=\sqrt{15}$ hay $x=-\sqrt{15}$ hay $x=\sqrt{10}$ hay $x=-\sqrt{10}$