Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
y = \dfrac{1}{{{x^4} - 2{x^2} + 3}}\\
DKXD:\,\,\,{x^4} - 2{x^2} + 3 \ne 0\\
\Leftrightarrow \left( {{x^4} - 2{x^2} + 1} \right) + 2 \ne 0\\
\Leftrightarrow {\left( {{x^2} - 1} \right)^2} + 2 \ne 0,\,\,\,\forall x\\
\Rightarrow TXD:\,\,\,\,D = R\\
b,\\
y = \sqrt {x + 3 - 2\sqrt {x + 2} } \\
DKXD:\,\,\,\left\{ \begin{array}{l}
x + 2 \ge 0\\
x + 3 - 2\sqrt {x + 2} \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 2\\
\left( {x + 2} \right) - 2\sqrt {x + 2} + 1 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 2\\
{\left( {\sqrt {x + 2} - 1} \right)^2} \ge 0
\end{array} \right.\\
\Leftrightarrow x \ge - 2\\
\Rightarrow TXD:\,\,\,\,\,\,\,D = \left[ { - 2; + \infty } \right)\\
c,\\
y = \sqrt[3]{{{x^2} - 4}} + \sqrt {{x^2} - 4x + 4} \\
DKXD:\,\,\,\,{x^2} - 4x + 4 \ge 0 \Leftrightarrow {\left( {x - 2} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow TXD:\,\,\,\,D = R
\end{array}\)
