Предмет: Алгебра, автор: 9Delight6

Найдите область определения функции

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Ответы

Автор ответа: Аноним

Подкоренное выражение должно быть больше нуля, т.к. стоит в знаменателе и под корнем. т.е. х²-6х+9>0, т.к. х²-6х+9=(х-3)²≥0 при любом действительном х, то х≠3

Значит, область определения - все числа, кроме 3

Ответ х∈(-∞;3)∪(3;+∞)

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