Đk: \(x\ge-2\)
\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\)
\(\Leftrightarrow\left|\sqrt{x+2}-2\right|+\left|\sqrt{x+2}-3\right|=1\) (*)
TH1: \(\sqrt{x+2}-3\ge0\)
(*) \(\Leftrightarrow\sqrt{x+2}-2+\sqrt{x+2}-3=1\)
\(\Leftrightarrow2\sqrt{x+2}=6\Leftrightarrow\sqrt{x+2}=3\Leftrightarrow x+2=9\Leftrightarrow x=7\left(N\right)\)
TH2: \(\sqrt{x+2}-2< 0\)
(*) \(\Leftrightarrow-\sqrt{x+2}+2-\sqrt{x+2}+3=1\)
\(\Leftrightarrow-2\sqrt{x+2}=-4\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(L\right)\)
TH3: \(\left\{{}\begin{matrix}\sqrt{x+2}-2\ge0\\\sqrt{x+2}-3< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+2}\ge2\\\sqrt{x+2}< 3\end{matrix}\right.\) \(\Leftrightarrow2\le\sqrt{x+2}< 3\) \(\Leftrightarrow4\le x+2< 9\) \(\Leftrightarrow2\le x< 7\)
(*) \(\Leftrightarrow1=1\) (luôn đúng)
Kl: 2\< x \< 7
