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