Giải thích các bước giải:
ĐKXĐ : $-2\le x\le 2$
Ta có :
$6+2\sqrt{4-x^2}=3(\sqrt{2+x}+\sqrt{2-x})$
$\to 2+(2+x)+(2-x)+2\sqrt{2+x}.\sqrt{2-x}=3(\sqrt{2+x}+\sqrt{2-x})$
$\to 2+(\sqrt{2+x}+\sqrt{2-x})^2=3(\sqrt{2+x}+\sqrt{2-x})$
$\to (\sqrt{2+x}+\sqrt{2-x})^2-3(\sqrt{2+x}+\sqrt{2-x})+2=0$
$\to (\sqrt{2+x}+\sqrt{2-x}-2)(\sqrt{2+x}+\sqrt{2-x}-1)=0$
$+)\sqrt{2+x}+\sqrt{2-x}-2=0$
$\to\sqrt{2+x}+\sqrt{2-x}=2$
Mà $\sqrt{2+x}+\sqrt{2-x}\ge \sqrt{2+x+2-x}=2$
Dấu = xảy ra khi $(2+x)(2-x)=0\to x\in\{2,-2\}$
$+)\sqrt{2+x}+\sqrt{2-x}-1=0$
$\to \sqrt{2+x}+\sqrt{2-x}=1$ vô nghiệm vì $\sqrt{2+x}+\sqrt{2-x}\ge 2$ (cmt)
