1)б,в
4)ОДЗ: x>0
![[tex]log_{0.1}(x^2-8)-log_{0.1}2x=0\\log_{0.1}(\frac{x^2-8}{2x})=0\\\frac{x^2-8}{2x}=(0.1)^0\ \ \ \ \ \ |*2x\neq0\\x^2-8=2x\\x^2-2x-8=0\\x_1=4\ \ \ \ \ \ x_2=-2 [tex]log_{0.1}(x^2-8)-log_{0.1}2x=0\\log_{0.1}(\frac{x^2-8}{2x})=0\\\frac{x^2-8}{2x}=(0.1)^0\ \ \ \ \ \ |*2x\neq0\\x^2-8=2x\\x^2-2x-8=0\\x_1=4\ \ \ \ \ \ x_2=-2](https://tex.z-dn.net/?f=%5Btex%5Dlog_%7B0.1%7D%28x%5E2-8%29-log_%7B0.1%7D2x%3D0%5C%5Clog_%7B0.1%7D%28%5Cfrac%7Bx%5E2-8%7D%7B2x%7D%29%3D0%5C%5C%5Cfrac%7Bx%5E2-8%7D%7B2x%7D%3D%280.1%29%5E0%5C+%5C+%5C+%5C+%5C+%5C+%7C%2A2x%5Cneq0%5C%5Cx%5E2-8%3D2x%5C%5Cx%5E2-2x-8%3D0%5C%5Cx_1%3D4%5C+%5C+%5C+%5C+%5C+%5C+x_2%3D-2)
Ответ:x=4
5)
1\\\frac{4x-5}{2}\leq-2\\4x-5\leq-4\\4x\leq1\\x\leq\frac{1}{4}" alt="(\sqrt{6})^{4x-5}\leq\frac{1}{36}\\6^{\frac{4x-5}{2}}\leq6^{-2}\ \ \ \ \ \ \ \ \ 6>1\\\frac{4x-5}{2}\leq-2\\4x-5\leq-4\\4x\leq1\\x\leq\frac{1}{4}" align="absmiddle" class="latex-formula">
3)Функция не является ни чётной ни нечётной, т.к.
и 