Đáp án:
$\begin{array}{l}
a)\dfrac{{4x}}{{{x^2} - 5x + 6}}\\
Dkxd:{x^2} - 5x + 6 \ne 0\\
\Leftrightarrow {x^2} - 2x - 3x + 6 \ne 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x - 3} \right) \ne 0\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ne 2\\
x \ne 3
\end{array} \right.\\
Vậy\,x \ne 2;x \ne 3\\
b)\dfrac{{6x}}{{{x^2} - x + 1}}\\
Dkxd:{x^2} - x + 1 \ne 0\\
\Leftrightarrow {x^2} - 2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4} \ne 0\\
\Leftrightarrow {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ne 0\left( {tm} \right)\\
Vậy\,x \in R\\
c)\dfrac{7}{{4{x^2} - 12x + 9}}\\
Dkxd:4{x^2} - 12x + 9 \ne 0\\
\Leftrightarrow {\left( {2x - 3} \right)^2} \ne 0\\
\Leftrightarrow x \ne \dfrac{3}{2}\\
Vậy\,x \ne \dfrac{3}{2}
\end{array}$
