Giải thích các bước giải:

Các hàm số đã cho xác định khi và chỉ khi:

\(\begin{array}{l}
o,\\
y = \sqrt {x + 2\sqrt {x - 1} } \\
DKXD:\,\,\,\left\{ \begin{array}{l}
x - 1 \ge 0\\
x + 2\sqrt {x - 1}  \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left( {x - 1} \right) + 2\sqrt {x - 1}  + 1 \ge 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
{\left( {\sqrt {x - 1}  + 1} \right)^2} \ge 0
\end{array} \right. \Leftrightarrow x \ge 1\\
 \Rightarrow TXD:\,\,\,\,\,\,D = \left[ {1; + \infty } \right)\\
p,\\
DKXD:\,\,\,\,\,\left\{ \begin{array}{l}
x + \dfrac{1}{4} \ge 0\\
x + \dfrac{1}{2} + \sqrt {x + \dfrac{1}{4}}  \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{1}{4}\\
\left( {x + \dfrac{1}{4}} \right) + \sqrt {x + \dfrac{1}{4}}  + \dfrac{1}{4} \ge 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{1}{4}\\
{\left( {\sqrt {x + \dfrac{1}{4}}  + \dfrac{1}{2}} \right)^2} \ge 0
\end{array} \right. \Leftrightarrow x \ge  - \dfrac{1}{4}\\
 \Rightarrow TXD:\,\,\,\,\,D = \left[ { - \dfrac{1}{4}; + \infty } \right)\\
r,\\
DKXD:\,\,\,\,\left| {2x - 4} \right| + \left| {{x^2} + 2x} \right| \ne 0,\,\,\,\,\forall x\\
 \Rightarrow TXD:\,\,\,\,D = R\\
s,\\
y = \dfrac{{\sqrt {2x - 1} }}{{\left| {x + 1} \right| - 2}}\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
2x - 1 \ge 0\\
\left| {x + 1} \right| - 2 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
x + 1 \ne 2\\
x + 1 \ne  - 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
x \ne 1
\end{array} \right.\\
 \Rightarrow TXD:\,\,\,\,\,D = \left[ {\dfrac{1}{2}; + \infty } \right)\backslash \left\{ 1 \right\}
\end{array}\)