Odpowiedź :
[tex]D:2x-1>0\\D:2x>1\\D:x>\dfrac{1}{2}\\\\\log_{0,1}(2x-1)\geq0\\\log_{0,1}(2x-1)\geq\log_{0,1}1\\2x-1\leq1\\2x\leq2\\x\leq1\\\\x\leq1\wedge x>\dfrac{1}{2}\\\boxed{x\in\left(\dfrac{1}{2},1\right\rangle}[/tex]
[tex]D:2x-1>0\\D:2x>1\\D:x>\dfrac{1}{2}\\\\\log_{0,1}(2x-1)\geq0\\\log_{0,1}(2x-1)\geq\log_{0,1}1\\2x-1\leq1\\2x\leq2\\x\leq1\\\\x\leq1\wedge x>\dfrac{1}{2}\\\boxed{x\in\left(\dfrac{1}{2},1\right\rangle}[/tex]