Đáp án:
$\begin{array}{l}
e)y = \dfrac{{4 - 5x}}{{{x^3} - {x^2} + x - 1}}\\
Dkxd:{x^3} - {x^2} + x - 1 \ne 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {{x^2} + 1} \right) \ne 0\\
\Leftrightarrow x - 1 \ne 0\left( {do:{x^2} + 1 > 0} \right)\\
\Leftrightarrow x \ne 1\\
Vậy\,TXD:D = R\backslash \left\{ 1 \right\}\\
f)y = \sqrt {\dfrac{{x + 1}}{{{x^2} + 1}}} \\
Dkxd:\dfrac{{x + 1}}{{{x^2} + 1}} \ge 0\\
\Leftrightarrow x + 1 \ge 0\left( {do:{x^2} + 1 > 0} \right)\\
\Leftrightarrow x \ge - 1\\
Vậy\,TXD:D = \left[ { - 1; + \infty } \right)\\
g)y = \sqrt {{x^2} - 4} \\
Dkxd:{x^2} - 4 \ge 0\\
\Leftrightarrow {x^2} \ge 4\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 2\\
x \le - 2
\end{array} \right.\\
Vậy\,TXD:D = R\backslash \left( { - 2;2} \right) = \left( { - \infty ; - 2} \right] \cup \left[ {2; + \infty } \right)
\end{array}$
