найти область определения фунции
[tex]y = \frac{4( {x}^{2} - 16) }{ \sqrt{ {x}^{2} - 8x + 7} } [/tex]
[tex]y = \frac{4( {x}^{2} - 16) }{ \sqrt{ {x}^{2} - 8x + 7} } [/tex]
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Ответ:
Объяснение:
x^2 -8x+7>0, x=1, x=7 + + + + + +(1) - - - - - - - - (7) + + + + +
D(y) : (-Б;1) u (7; +Б)