Предмет: Математика, автор: goroshekbb

Найдите область определения функции
y=1+ 2/3х-1

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Автор ответа: yevheniiavz

0

$y=1+\frac{2}{3x-1}\\3x-1\neq 0\\3x\neq 1\\x\neq \frac{1}{3} \\D(y)=(-\infty;\;\frac{1}{3} )\cup(\frac{1}{3} ;\;+\infty).$

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