$g'(x)=1-\dfrac{2x+2}{2\sqrt{x^2+2x-3}}$
$g'(x)\ge 0$
$\Leftrightarrow \dfrac{2x+2}{2\sqrt{x^2+2x-3}}-1\le 0$
$\Leftrightarrow \dfrac{2x+2-2\sqrt{x^2+2x-3}}{2\sqrt{x^2+2x-3}}\le 0$
ĐK: $\sqrt{x^2+2x-3}\ne 0$
Ta có $2\sqrt{x^2+2x-3}>0$
$\Rightarrow 2x+2-2\sqrt{x^2+2x-3}\le 0$
$\Leftrightarrow \sqrt{x^2+2x-3}\ge x+1$ (*)
+ Nếu $x+1<0\Leftrightarrow x<-1$ $(*)\Rightarrow x^2+2x-3>0$ (theo ĐK)
$\Leftrightarrow x<-3$ hoặc $x>1$
$\Rightarrow x<-3$
+ Nếu $x+1\ge 0\Leftrightarrow x\ge -1$
$(*) \Rightarrow x^2+2x-3\ge x^2+2x+1$ (vô lí)
Vậy $x<-3$
