Объяснение:

[tex]\displaystyle\\\sqrt{3-2x}+\frac{x^2-1}{\sqrt{x-1} } .\\\\[/tex]

ОДЗ:

[tex]\displaystyle\\\left \{ {{3-2x\geq 0} \atop {x-1 > 0}} \right. \ \ \ \ \ \ \left \{ {{3\geq 2x\ |:2} \atop {x > 1}} \right. \ \ \ \ \ \ \left \{ {{x\leq 1,5} \atop {x > 1}} \right. \ \ \ \ \ \ \Rightarrow\\\\[/tex]

Ответ: x∈(1;1,5].