Đáp án:
$\begin{array}{l}
a)y = \frac{{\sqrt {x + 2} }}{{2x + 3}} - \sqrt 2 - 1\\
Xđ:\left\{ \begin{array}{l}
x + 2 \ge 0\\
2x + 3 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge - 2\\
x \ne \frac{{ - 3}}{2}
\end{array} \right.\\
b)y = \frac{{ - 4x + 3}}{{\sqrt {x + 1} - \sqrt {2x - 5} }}\\
Xđ:\left\{ \begin{array}{l}
x + 1 \ge 0\\
2x - 5 \ge 0\\
\sqrt {x + 1} - \sqrt {2x - 5} \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge - 1\\
x \ge \frac{5}{2}\\
\sqrt {x + 1} \ne \sqrt {2x - 5}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge \frac{5}{2}\\
x + 1 \ne 2x - 5
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge \frac{5}{2}\\
x \ne 6
\end{array} \right.\\
c)y = \frac{{\sqrt {2x + 1} }}{{x - 1}} - \sqrt 2 - 1\\
Xác định:\left\{ \begin{array}{l}
2x + 1 \ge 0\\
x - 1 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge \frac{{ - 1}}{2}\\
x \ne 1
\end{array} \right.
\end{array}$
