116. Find the minimum value of the fraction $\frac{x^{2}-3 x+3}{1-x}$, if $x<1$.
Solution. I method. The given fraction can be represented as $\frac{x^{2}}{1-x}+3$. For $x<1$, the first term is non-negative and equals 0 only when $x=0$. Since 0 is included in the set of values for $x$, the given fraction achieves its minimum value of 3 when $x=0$.

II method. Representing the given fraction as the sum

$$
1-x+\frac{1}{1-x}+1
$$

and using the fact that the sum of two positive reciprocals is not less than two, we get that the given fraction is not less than 3, i.e., its minimum value is 3.

Answer: 3.

$10^{*}$