0\\\\x^{log_{5}x }=(5^{log_{5}x })^{log_{5}x }=5^{log_{5}^{2}x}\\\\5^{log_{5}^{2}x }+x^{log_{5}x }\geq2\sqrt[4]{5} \\\\5^{log_{5}^{2}x}+5^{log_{5}^{2}x}\geq 2\sqrt[4]{5}\\\\2*5^{log_{5}^{2}x}\geq 2\sqrt[4]{5}\\\\5^{log_{5}^{2}x}\geq5^{\frac{1}{4} }\\\\log_{5}^{2} x\geq \frac{1}{4}" alt="1)x>0\\\\x^{log_{5}x }=(5^{log_{5}x })^{log_{5}x }=5^{log_{5}^{2}x}\\\\5^{log_{5}^{2}x }+x^{log_{5}x }\geq2\sqrt[4]{5} \\\\5^{log_{5}^{2}x}+5^{log_{5}^{2}x}\geq 2\sqrt[4]{5}\\\\2*5^{log_{5}^{2}x}\geq 2\sqrt[4]{5}\\\\5^{log_{5}^{2}x}\geq5^{\frac{1}{4} }\\\\log_{5}^{2} x\geq \frac{1}{4}" align="absmiddle" class="latex-formula">
3log_{3}x\\\\log_{3}x=m\\\\m^{2}-3m+2>0\\\\(m-1)(m-2)>0\\\\m<1;m>2\\\\log_{3}x<1\\\\x<3\\\\log_{3}x>2\\\\x>9\\\\x\in(0;3)\cup(9;+\infty)" alt="2)log_{3}^{2}x+2>3log_{3}x\\\\log_{3}x=m\\\\m^{2}-3m+2>0\\\\(m-1)(m-2)>0\\\\m<1;m>2\\\\log_{3}x<1\\\\x<3\\\\log_{3}x>2\\\\x>9\\\\x\in(0;3)\cup(9;+\infty)" align="absmiddle" class="latex-formula">
Ответ :