a, ĐKXĐ: \(-1\le x\le6\)
\(pt\Leftrightarrow1+x+1+2\sqrt{x+1}=6-x\)
\(\Leftrightarrow\sqrt{x+1}=2-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=\left(2-x\right)^2\\2-x\ge0\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{5-\sqrt{13}}{2}\left(tm\right)\)
b, ĐKXĐ: \(3-\sqrt{3}\le x\le3+\sqrt{3}\)
\(pt\Leftrightarrow x^2-6x+6-4\sqrt{x^2-6x+6}+3=0\)
\(\Leftrightarrow\left(\sqrt{x^2-6x+6}-1\right)\left(\sqrt{x^2-6x+6}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=1\\\sqrt{x^2-6x+6}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x+6=1\\x^2-6x+6=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)\left(x-1\right)=0\\x=3\pm2\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(tm\right)\\x=3\pm2\sqrt{3}\left(tm\right)\end{matrix}\right.\)