Ответ:
[tex](-\infty;-4)\cup(-4;-3)\cup(3;4)\cup)(4;+\infty)[/tex]
Объяснение:
[tex]1.\\\\x^2-9 > 0\\\\(x+3)(x-3) > 0\\\\x\in(-\infty;-3)\cup(3;+\infty)[/tex]
[tex]2.\\\\|x|-4=0\\\\|x|=4\\\\x=-4\ \ \ \ \ \ \ \ \ \ x=4\\\\x\in R\setminus\{-4;4\}[/tex]
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[tex]x\in (-\infty;-4)\cup(-4;-3)\cup(3;4)\cup)(4;+\infty)[/tex]
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Answers & Comments
Ответ:
[tex](-\infty;-4)\cup(-4;-3)\cup(3;4)\cup)(4;+\infty)[/tex]
Объяснение:
[tex]1.\\\\x^2-9 > 0\\\\(x+3)(x-3) > 0\\\\x\in(-\infty;-3)\cup(3;+\infty)[/tex]
[tex]2.\\\\|x|-4=0\\\\|x|=4\\\\x=-4\ \ \ \ \ \ \ \ \ \ x=4\\\\x\in R\setminus\{-4;4\}[/tex]
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[tex]x\in (-\infty;-4)\cup(-4;-3)\cup(3;4)\cup)(4;+\infty)[/tex]