$2.1 \quad \frac{2}{3-x}+\frac{1}{2}=\frac{6}{x(3-x)}$.
2.1 Let's move all terms of the equation to the left side and transform the obtained equation to the form

$$
\frac{x^{2}-7 x+12}{2 x(3-x)}=0
$$

From the equation $x^{2}-7 x+12=0$, we find $x_{1}=3, x_{2}=4$. When $x=3$, the denominator becomes zero; therefore, 3 is not a root. Thus, $x=4$.

Answer: $x=4$.