найти область определения фунции
[tex]y = \frac{1}{ \sqrt{144 - 9 {x}^{2} } } [/tex]
[tex]y = \frac{1}{ \sqrt{144 - 9 {x}^{2} } } [/tex]
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Ответ:
Объяснение:
144-9x^2>0, (12-3x)(12+3x)>0 x=4 u x=-4 - - - - - - - (-4) + + + + + + (4) - - - - -
D(y) : (-4;4)