Question

Помогите решить это ур-ние

Answer 1

$\frac{x^2-x-12}{x+3} = 0 \\ \left \{ {{ \frac{(x+3)(x-4)}{x+3} = 0} \atop {x \ne -3}} \right. \\ \left \{ {{x=4} \atop {x \ne -3}} \right. \\ x = 4$

Answer 2

$\frac{ x^{2}-x-12}{x+3} =0$

$x+3 \neq 0 =\ \textgreater \ x\neq -3$

$x^{2} -x-12=0$
$\left \{ {{x_{1}+x_{2} =1} \atop {x_{1}x_{2} =-12}} \right. =\ \textgreater \ \left \{ {{x_{1} =4} \atop {x_{2}=-3}} \right.$
$x=4$

Ответ: 4