Question

Решить неравенство:
log2(4x^2-1)-log2(x)≤log2(5x+9/x-11)

Answer 1

$log_{2}(4x^2-1)-log_{2}(x) leq log_{2}(5x+ frac{9}{x} -11) \ \O DZ: \ \ 4x^2-1 textgreater 0 \ x textgreater 0 \ 5x+ frac{9}{x} -11 textgreater 0 \ x neq 0 \ \ (2x-1)(2x+1) textgreater 0 \ x textgreater 0 \ 5x^2-11x+9 textgreater 0;D=121-4*45=121-180=-35 textless 0; \ \ xin(-oo;-1/2)U(1/2;+oo) \ xn(0;+oo) \ \ ODZ:xin(1/2;+oo)$
2>1, знак сохраняем

$frac{4x^2-1}{x} leq 5x+ frac{9}{x} -11 \ \ frac{4x^2-1}{x} leq frac{5x^2-11x+9}{x} \ \ frac{5x^2-11x+9-4x^2+1}{x} geq 0 \ \ frac{x^2-11x+10}{x} geq 0 \ \ D=121-40=81=9^2; x_{1} =10; x_{2} =1 \ \ frac{(x-10)(x-1)}{x} geq 0 \ \ ---(0)+++[1]---[10]++++ \ \ xin(0;1]U[10;+oo)$

с учетом ОДЗ,получаем ответ:
$xin( frac{1}{2} ;1]U[10;+oo)$