Đáp án: $x\in\{-4,2,-1\pm 2\sqrt{6}\}$
Giải thích các bước giải:
$(x+3)(1-x)=-5\sqrt{x^2+2x-7}$
$\rightarrow (x-1)(x+3)-5\sqrt{x^2+2x-7}=0$
$\rightarrow x^2+2x-3-5\sqrt{x^2+2x-7}=0$
$\rightarrow x^2+2x-7-5\sqrt{x^2+2x-7}+4=0$
$\rightarrow (\sqrt{x^2+2x-7}-1)(\sqrt{x^2+2x-7}-4)=0$
$\rightarrow \sqrt{x^2+2x-7}-1=0\rightarrow x^2+2x-8=0\rightarrow x\in\{-4,2\}$
Hoặc $\sqrt{x^2+2x-7}=4\rightarrow x^2+2x-23=0\rightarrow x=-1\pm 2\sqrt{6}$
