Неравенства с корнями, задания(3) в закрепе

Решите задачу:

$\sqrt{2x-1}\ \textgreater \ 2x-3\\\\1)\quad \left \{ {{2x-3 \ \textgreater \ 0} \atop {2x-1\ \textgreater \ 4x^2-12x+9}} \right. \; \left \{ {{x\ \textgreater \ \frac{3}{2}} \atop {4x^2-14x+10\ \textless \ 0}} \right. \; \left \{ {{x\ \textgreater \ 1,5} \atop {2x^2-7x+5\ \textless \ 0}} \right. \\\\2x^2-7x+5=0\; ,\; \; D=49-40=9,\; x_1=2,5\; ;\; x_2=1\\\\+++(1)---(2,5)+++\\\\x\in (1;\, 2,5)\; \; \; \left \{ {{x\ \textgreater \ 1,5} \atop {x\in (1\, ;\; 2,5)}} \right. \; \to \; \; x\in (1,5\; ;\; 2,5)\\\\ili$

$2)\; \; \left \{ {{2x-1\ \textgreater \ 0} \atop {2x-3\ \textless \ 0}} \right. \; \left \{ {{x\ \textgreater \ \frac{1}{2}} \atop {x\ \textless \ \frac{3}{2}}} \right.\; \left \{ {{x\ \textless \ 1,5} \atop {x\ \textgreater \ 0,5}} \right. \; \to \; x\in (0,5\; ;\; 1,5)$

$Otvet:\; \; x\in (0,5\, ;\, 1,5)\cup (1,5\; ;2,5)$