Đáp án:

\(\begin{array}{l}
1,\\
x = 4\\
2,\\
x =  - 1\\
3,\\
x = 4\\
4,\\
x = 1\\
5,\\
x = 3\\
6,\\
x = 2
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
DKXD:\,\,\,x \ge  - 5\\
\sqrt {x + 5}  = 3\\
 \Leftrightarrow x + 5 = {3^2}\\
 \Leftrightarrow x + 5 = 9\\
 \Leftrightarrow x = 4\\
2,\\
DKXD:\,\,\,x \ge  - 2\\
1 - \sqrt {x + 2}  = 0\\
 \Leftrightarrow \sqrt {x + 2}  = 1\\
 \Leftrightarrow x + 2 = {1^2}\\
 \Leftrightarrow x =  - 1\\
3,\\
DKXD:\,\,\,x \ge \dfrac{5}{2}\\
\sqrt {2x - 5}  - \sqrt {x - 1}  = 0\\
 \Leftrightarrow \sqrt {2x - 5}  = \sqrt {x - 1} \\
 \Leftrightarrow 2x - 5 = x - 1\\
 \Leftrightarrow 2x - x =  - 1 + 5\\
 \Leftrightarrow x = 4\\
4,\\
DKXD:\,\,\,x \ge  - 1\\
\sqrt {4x + 4}  = \sqrt 8 \\
 \Leftrightarrow 4x + 4 = 8\\
 \Leftrightarrow 4x = 4\\
 \Leftrightarrow x = 1\\
5,\\
DKXD:\,\,\,x \ge 2\\
\sqrt {x - 2} .\sqrt {x + 1}  = 2\\
 \Leftrightarrow \sqrt {\left( {x - 2} \right)\left( {x + 1} \right)}  = 2\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x + 1} \right) = {2^2}\\
 \Leftrightarrow {x^2} - x - 2 = 4\\
 \Leftrightarrow {x^2} - x - 6 = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x =  - 2
\end{array} \right.\\
x \ge 2 \Rightarrow x = 3\\
6,\\
DKXD:\,\,\,x \ge 2\\
\sqrt {{x^2} - 4}  - \sqrt {x - 2}  = 0\\
 \Leftrightarrow \sqrt {\left( {x - 2} \right)\left( {x + 2} \right)}  - \sqrt {x - 2}  = 0\\
 \Leftrightarrow \sqrt {x - 2} .\left( {\sqrt {x + 2}  - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 2}  = 0\\
\sqrt {x + 2}  - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 2}  = 0\\
\sqrt {x + 2}  = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 2 = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
x \ge 2 \Rightarrow x = 2
\end{array}\)