ĐK: \(\left\{{}\begin{matrix}3-x\ge0\left(1\right)\\9-x^2\ge0\left(2\right)\\24+8\sqrt{9-x^2}\ge0\left(3\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x\le3\)
(2) \(\Leftrightarrow x^2\le9\Rightarrow-3\le x\le3\)
(3) \(\Leftrightarrow4\left(3+x\right)+8\sqrt{\left(3-x\right)\left(3+x\right)}+4\left(3-x\right)\ge0\)
\(\Leftrightarrow\left(2\sqrt{3+x}+2\sqrt{3-x}\right)^2\ge0\) (luôn đúng)
Vậy đk của PT là \(-3\le x\le3\)
Ta có:
\(\sqrt{24+8\sqrt{9-x^2}}=x+2\sqrt{3-x}+4\)
\(\Rightarrow\sqrt{\left(2\sqrt{3+x}+2\sqrt{3-x}\right)^2}=x+2\sqrt{3-x}+4\)
\(\Rightarrow2\sqrt{3+x}+2\sqrt{3-x}=x+2\sqrt{3-x}+4\)
\(\Rightarrow x-2\sqrt{3+x}+4=0\\ \Rightarrow\left(x+3\right)-2\sqrt{3+x}+1=0\\ \Rightarrow\left(\sqrt{x+3}-1\right)^2=0\\ \Rightarrow\sqrt{x+3}=1\\ \Rightarrow x+3=1\\ \Rightarrow x=-2\left(TM\right)\)
