Question

СРОЧНО!!!!!!!!!$\frac{x^{2}-16 }{x^{2}-8x+16}=0\\\frac{3}{x^{2} -9}+\frac{x}{x-3}=1$

Answer 1

Ответ:

$\frac{x {}^{2} - 16}{ x {}^ {2} - 8 + 16} = 0 = x = - 4$

$\frac{3}{x {}^{2} - 9} + \frac{x}{x - 3 } = 1 = x = - 4$