Question

Помогите пожалуйста!

Answer 1

$1)x>0\\x^{log_{5}x }=(5^{log_{5}x })^{log_{5}x }=5^{log_{5}^{2}x}\\5^{log_{5}^{2}x }+x^{log_{5}x }geq2sqrt[4]{5} \\5^{log_{5}^{2}x}+5^{log_{5}^{2}x}geq 2sqrt[4]{5}\\2*5^{log_{5}^{2}x}geq 2sqrt[4]{5}\\5^{log_{5}^{2}x}geq5^{frac{1}{4} }\\log_{5}^{2} xgeq frac{1}{4}$

$(log_{5}x-frac{1}{2})(log_{5}+frac{1}{2})geq0\\(log_{5}x-log_{5}5^{frac{1}{2} })(log_{5}x-log_{5}5^{-frac{1}{2} })geq0\\(5-1)(x-sqrt{5})(5-1)(x-frac{1}{sqrt{5} })geq0\\(x-sqrt{5})(x-frac{1}{sqrt{5} })geq0\\xin(0;frac{1}{sqrt{5} }]cup[sqrt{5};+ infty)$

$2)log_{3}^{2}x+2>3log_{3}x\\log_{3}x=m\\m^{2}-3m+2>0\\(m-1)(m-2)>0\\m<1;m>2\\log_{3}x<1\\x<3\\log_{3}x>2\\x>9\\xin(0;3)cup(9;+infty)$

Ответ :

$xin(0;frac{1}{sqrt{5} }]cup[sqrt{5};3)cup(9;+infty)$