Question

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

Answer 1

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

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

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