Question

помогите срочно!Математика

Answer 1

$frac{x^2}{x-4}+frac{16}{4-x}=0\O.D.3.:\begin{cases}x-4neq0\4-xneq0end{cases}Rightarrow xneq4\\frac{x^2}{x-4}-frac{16}{x-4}=0\frac{x^2-16}{x-4}=0\x^2-16=0\x^2=16\x=pm4\no;O.D.3.;xneq4Rightarrow x=-4$