Question

Ответ: -1; 3
решите ​

Answer 1

ОДЗ :

$\left \{ {{x+2>0 \ ; \ 2x+7>0} \atop {2x+7\neq1 }} \right.\\\\\left \{ {{x>-2 \ ; \ x>-3,5} \atop {x\neq-3 }} \right. \ \Rightarrow \ x\in(-2 \ ; \ -\infty)\\\\\\log_{2x+7}(x+2)=\frac{1}{2} \\\\\sqrt{2x+7}=x+2\\\\(\sqrt{2x+7})^{2} =(x+2)^{2}\\\\2x+7=x^{2}+4x+4\\\\x^{2} +2x-3=0\\\\x_{1} +x_{2}=-2\\\\x_{1} *x_{2} =-3-teorema \ Vieta\\\\x_{1} =-3-neyd\\\\x_{2}=1 \\\\Otvet:\boxed{x=1}$