\[\boxed{\text{276\ (276).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[1)\ y = \frac{6}{3 - x};\ \ \ \ возрастает\ \]
\[на\ (3;\ + \infty),\ так\ как:\]
\[x_{1} = 5:\ \]
\[y_{1} = \frac{6}{- 2} = - 3.\]
\[x_{2} = 6:\ \ \]
\[y_{2} = \frac{6}{- 3} = - 2.\]
\[x_{2} > x_{1};\ \ y_{2} > y_{1}.\]
\[2)\ y = x^{2} - 4x + 3,\ убывает\ \]
\[на\ ( - \infty;2\rbrack,\ так\ как:\]
\[x_{1} = 0:\ \ \]
\[y_{1} = 0 - 0 + 3 = 3.\]
\[x_{2} = 1:\ \ \]
\[y_{2} = 1 - 4 + 3 = 0.\]
\(x_{2} > x_{1};\text{\ \ }y_{2} < y_{1}\text{.\ }\)