Найдите область определения функции: [tex]y= \frac{ \sqrt{16- x^{2} } }{x+3} [/tex]. Created by kopylovaulyana. algebra-ru

tex]...

July 2022 2 7 Report

Найдите область определения функции:
[tex]y= \frac{ \sqrt{16- x^{2} } }{x+3} [/tex]

mukus13

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$y= \frac{ \sqrt{16-x^2} }{x+3}$

$\left \{ {{16-x^2 \geq 0} \atop {x+3 \neq 0}} \right.$

$\left \{ {{(4-x)(4+x)\geq 0} \atop {x \neq -3}} \right.$

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Ответ: $[-4;-3)$ ∪ $(-3;4]$

14 votes Thanks 12

kopylovaulyana огромное спасибо))

kopylovaulyana а можешь еще два решить?

kisonix

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1 votes Thanks 1

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