\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\) ( ĐK : \(x\ge-2\) )
\(\Leftrightarrow\sqrt{x+2-4\sqrt{x+2}+4}+\sqrt{x+2-6\sqrt{x+2}+9}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+2}-2\right)^2}+\sqrt{\left(\sqrt{x+2}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x+2}-2\right|+\left|\sqrt{x+2}-3\right|=1\)
Ta có : \(\left|\sqrt{x+2}-2\right|+\left|\sqrt{x+2}-3\right|=\left|\sqrt{x+2}-2\right|+\left|3-\sqrt{x+2}\right|\)
Áp dụng BĐT : \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
\(\Rightarrow\left|\sqrt{x+2}-2\right|+\left|3-\sqrt{x+2}\right|\ge\left|\sqrt{x+2}-2+3-\sqrt{x+2}\right|=1\)
Dấu \("="\) xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x+2}-2\ge0\\3-\sqrt{x+2}\ge0\end{matrix}\right.\Leftrightarrow2\le x\le7\)
