Log0,5(4+x)/(x-1)>=2

$log_{0,5} \frac{4+x}{x-1} \geq 2 2= log_{0,5} 0,5^{2} = log_{0,5} 0,25$
$log_{0,5} \frac{4+x}{x-1} \geq log_{0,5} 0,25$
основание логарифма а=0,5. 0<0,5<1<br> знак неравенства меняем:
$\left \{ {{ \frac{4+x}{x-1} \ \textgreater \ 0} \atop { \frac{4+x}{x-1} \leq 0,25 }} \right. \left \{ {{ \frac{4+x}{x-1} \ \textgreater \ 0} \atop { \frac{4+x}{x-1}-0,25 \leq 0 }} \right.$
$1. \frac{4+x}{x-1} \ \textgreater \ 0$ ,метод интервалов:
$\left \{ {{x+4=0} \atop {x-1 \neq 0}} \right. \left \{ {{x=-4} \atop {x \neq 1}} \right.$
+ - +
-----------(-4)-----------(1)----------------->x
x∈(-4;∞)∪(1;∞)

$2. \frac{4+x}{x-1}-0,25 \leq 0.$
$\frac{4+x-0,25x+0,25}{x-1} \leq 0, \frac{0,75x+4,25}{x-1} \leq 0$
метод интервалов:
$\left \{ {{0,75x+4,25=0} \atop {x-1 \neq 0}} \right. \left \{ {{x= -\frac{17}{3} } \atop {x \neq 1}} \right.$
+ - +
----------[-17/3]-------------(1)--------------->x

x∈[-17/3;1)
/ / / / / / / / / / / / / / / /
------------[-17/3]------(-4)-----------(1)------------------>x
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x∈[-17/3;-4)