Question

Решите пожалуйста. Пожалуйста

Answer 1

$\frac{1}{x^{2}-16}=\frac{5}{(x+4)^{2}}\\\\\frac{1}{(x+4)(x-4)} -\frac{5}{(x+4)^{2}}=0\\\\\frac{x+4-5(x-4)}{(x+4)^{2}(x-4)}=0\\\\\frac{x+4-5x+20}{(x+4)^{2}(x-4)}=0\\\\\frac{-4x+24}{(x+4)^{2}(x-4)}=0\\\\x\neq -4 ; x\neq 4\\\\-4x+24=0\\\\-4x=-24\\\\x=6$