Điều kiện \(x\ge1\)
\(\Rightarrow6+2x-2\sqrt{x-1}-4\sqrt{x+2}=0\)
\(\Leftrightarrow\left(x-1-2\sqrt{x-1}+1\right)+\left(x+2-4\sqrt{x+2}+4\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{x+2}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\x+2=4\end{matrix}\right.\)
\(\Leftrightarrow x=2\)