Question

Найди область определения функции :

Answer 1

$f(x)=frac{sqrt{2x-7} }{sqrt{3-2x}}\\left { {{2x-7geq0} atop {3-2x>0}} right.\\left { {{2xgeq7 } atop {-2x>-3}} right.\\left { {{xgeq3,5 } atop {x<1,5}} right.\\Otvet:boxed{xinoslash}$

Answer 2

$f(x)=frac{sqrt{2x-7} }{sqrt{3-2x}}$

$left { {{2x-7geq0} atop {3-2x>0}} right.$

$left { {{2xgeq7} atop {-2x>-3}} right.$

$left { {{xgeq3.5} atop {x<1.5}} right.$

------------------(1.5)------------------[3.5]-------------

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Ответ: ∅