Question

Найди область определения функции :

Answer 1

$f(x)=\frac{\sqrt{2x-7} }{\sqrt{3-2x}}\\\\\left \{ {{2x-7\geq0} \atop {3-2x>0}} \right.\\\\\left \{ {{2x\geq7 } \atop {-2x>-3}} \right.\\\\\left \{ {{x\geq3,5 } \atop {x<1,5}} \right.\\\\Otvet:\boxed{x\in\oslash}$

Answer 2

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

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

$\left \{ {{2x\geq7} \atop {-2x>-3}} \right.$

$\left \{ {{x\geq3.5} \atop {x<1.5}} \right.$

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

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Ответ: ∅