Question

помогите решить уравнениееее​

Answer 1

$\frac{x-4}{x+3}=\frac{2x+4}{x-3}\\\\\frac{x-4}{x+3}-\frac{2x+4}{x-3}=0\\\\\frac{(x-4)*(x-3)-(2x+4)*(x+3)}{(x+3)(x-3)}=0\\\\\frac{x^{2}-3x-4x+12-2x^{2}-6x-4x-12}{(x+3)(x-3)}=0\\\\\frac{-x^{2} -17x}{(x+3)(x-3)}=0\\\\\left \{ {{-x^{2}-17x=0 } \atop {x+3\neq0;x-3\neq0}} \right.\\\\\\\left \{ {{-x(x+17)=0} \atop {x\neq-3;x\neq3}} \right. \\\\\\\left \{ {{\left[\begin{array}{ccc}x_{1}=0 \\x_{2}=-17 \end{array}\right } \atop {x\neq-3;x\neq 3}} \right. \\\\Otvet:\boxed{0;-17}$