Question

Решить показательное неравенство

Answer 1

Ответ:

$9^{\frac{1}{x}-1}+2\cdot 3^{\frac{1}{x}-1}-3\geq 0\ \ ,\ \ \ \ \ ODZ:\ x\ne 0\ ,\\\\\Big(3^{\frac{1}{x}-1}\Big)^2+2\cdot 3^{\frac{1}{x}-1}-3\geq 0\\\\t=3^{\frac{1}{x}-1}>0\ \ ,\ \ \ \ t^2+2t-3\geq 0\ \ ,\ \ \ (t-1)(t+3)\geq 0\ \ ,\\\\znaki:\ \ \ +++[-3\, ]---[\, 1\, ]+++\ \ ,\ \ t\in (-\infty ;-3\ ]\cup [\ 1\ ;+\infty )\\\\t>0\ \ \to \ \ \ t\in [\ 1\ ;+\infty \, )\\\\3^{\frac{1}{x}-1}\geq 1\ \ ,\ \ \ \ 3^{\frac{1}{x}-1}\geq 3^0\ \ \ \Rightarrow \ \ \ \dfrac{1}{x}-1\geq 0\ \ ,\ \ \ \dfrac{1-x}{x}\geq 0\ ,$

$\dfrac{x-1}{x}\leq 0\ \ ,\ \ \ znaki:\ +++(\, 0\, )---[\ 1\ ]+++\\\\\\Otvet:\ x\in (\ 0\ ;\ 1\ ]\ .$