Объяснение:

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[tex]\displaystyle\bf\\ODZ: \ x-3 > 0 \ \ \ \Rightarrow \ \ \ x > 3\\\\2\log_{25}^{2} (x-3)-1 < \log_{25}(x-3)\\\\\log_{25} (x-3)=m\\\\2m^{2} -m-1 < 0\\\\2m^{2} -m-1=0\\\\D=(-1)^{2} -4\cdot 2\cdot(-1)=1+8=9=3^{2} \\\\\\m_{1} =\frac{1-3}{4}=-\frac{1}{2} \\\\\\m_{2} =\frac{1+3}{4} =1\\\\\\2m^{2} -m-1=2\cdot\Big(m+\frac{1}{2}\Big) \cdot\Big(m-1\Big)\\\\\Big(m+\frac{1}{2}\Big) \cdot\Big(m-1\Big) < 0\\\\\\+ + + + + \Big(-\frac{1}{2} \Big) - - - - - \Big(1\Big) + + + + +[/tex]

[tex]\displaystyle\bf\\\left \{ {{m > -\dfrac{1}{2} } \atop {m < 1}} \right. \\\\1)\\\\\log_{25} (x-3) > -\frac{1}{2} \\\\x-3 > 25^{-\frac{1}{2} } \\\\x-3 > 0,2\\\\x > 3,2\\\\2)\\\\\log_{25} (x-3) < 1\\\\x-3 < 25\\\\x < 28\\\\\\Otvet \ : \ x\in\Big(3,2 \ ; \ 28\Big)[/tex]