Решите уравнение x\x+4 + x+2\x-4 = 32\x^2-16

Решите задачу:

$\frac{x(x-4)+(x+2)(x+4)}{(x-4)(x+4)} = \frac{32}{ x^{2} -16} \\ \\ \frac{ x^{2} -4x+ x^{2} +4x+2x+8}{ x^{2} -16} =\frac{32}{ x^{2} -16} \\ \\ \frac{2 x^{2} +2x+8}{ x^{2} -16} = \frac{32}{ x^{2} -16} \\ \\ \frac{2 x^{2} +2x+8-32}{ x^{2} -16} =0 \\ \\ x1 \neq 4 \\ \\ x2 \neq -4 \\ \\ 2 x^{2} +2x-24=0 \\ \\ x^{2} +x-12=0 \\ \\ D=1+48=49 \\ \\ \sqrt{49} =7 \\ \\ x1=-1-7/2=-4 (ne- podxodit- po- usl )\\ \\ x2=-1+7/2=3 \\ \\ otvet:3$