Найдите область определения функции: y=корень из (x^2-18x+72).

\[y = \sqrt{x^{2} - 18x + 72}\]

\[x^{2} - 18x + 72 \geq 0\]

\[D = 81 - 72 = 9\]

\[x_{1} = 9 + 3 = 12;\ \ \ \]

\[x_{2} = 9 - 3 = 6.\]

\[(x - 6)(x - 12) \geq 0\]

\[D(y) = ( - \infty;6\rbrack \cup \lbrack 12; + \infty).\]