Đáp án:
$\begin{array}{l}
1)Dkxd:x \ge 2\\
\sqrt {x - 2} + \sqrt {4x - 8} = 0\\
\Leftrightarrow \sqrt {x - 2} + \sqrt {4\left( {x - 2} \right)} = 0\\
\Leftrightarrow \sqrt {x - 2} + 2\sqrt {x - 2} = 0\\
\Leftrightarrow \sqrt {x - 2} = 0\\
\Leftrightarrow x = 2\left( {tmdk} \right)\\
Vậy\,x = 2\\
2)Dkxd:x \ge 1\\
\sqrt {x - 1} - \sqrt {2x\left( {x - 1} \right)} = 0\\
\Leftrightarrow \sqrt {x - 1} .\left( {1 - \sqrt {2x} } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
\sqrt {2x} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\left( {tm} \right)\\
x = \dfrac{1}{2}\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 1\\
3)Dkxd:x \ge - 1\\
\sqrt {x + 1} + \sqrt {9x + 9} = 4\\
\Leftrightarrow \sqrt {x + 1} + 3\sqrt {x + 1} = 4\\
\Leftrightarrow \sqrt {x + 1} = 1\\
\Leftrightarrow x + 1 = 1\\
\Leftrightarrow x = 0\left( {tmdk} \right)\\
Vậy\,x = 0\\
4)Dkxd:x \ge 1\\
\sqrt {5x - 5} - \sqrt {35} = 0\\
\Leftrightarrow \sqrt 5 .\sqrt {x - 1} - \sqrt 5 .\sqrt 7 = 0\\
\Leftrightarrow \sqrt {x - 1} = \sqrt 7 \\
\Leftrightarrow x - 1 = 7\\
\Leftrightarrow x = 8\left( {tm} \right)\\
Vậy\,x = 8
\end{array}$
