Решить неравенство а)(х-2)log₀,₅х≤o b) xˣ∠1 ,x больше 0 с) lg(x-2)+lg(27-x)∠2

Question 1

Решить неравенство а)(х-2)log₀,₅х≤o
b) xˣ∠1 ,x больше 0
с) lg(x-2)+lg(27-x)∠2

Question 2

Условие пункта а) непонятно

Question 3

исправила

Question 4

Решите задачу:

$1)\; \; (x-2)log_{0,5}x \leq 0\; ,\; \; ODZ:\; \; x\ \textgreater \ 0\\\\ \left \{ {{x-2 \leq 0} \atop {log_{0,5}x \geq 0}} \right. \; \; \; ili\; \; \; \left \{ {{x-2 \geq 0} \atop {log_{0,5}x \leq 0}} \right. \\\\ \left \{ {{x \leq 2} \atop {0\ \textless \ x \leq 1}} \right. \; \; \; \; \; ili\; \; \; \; \; \left \{ {{x \geq 2} \atop {x \geq 1}} \right. \\\\0\ \textless \ x \leq 1\; \; \; \; ili\; \; \; \; \; x \geq 2\\\\Otvet;\; \; x\in (0,1\, ]\cup [\, 2,+\infty )$

$2)\; \; x^{x}\ \textless \ 1\; \; ,\; \; \; \; x\ \textgreater \ 0\\\\xlgx\ \textless \ lg1\\\\xlgx\ \textless \ 0\\\\ \left \{ {{x\ \textgreater \ 0} \atop {lgx\ \textless \ 0}} \right. \; \left \{ {{x\ \textgreater \ 0} \atop {x\ \textless \ 1}} \right. \\\\Otvet:\; \; x\in (0,1).$

$3)\; \; lg(x-2)+lg(27-x)\ \textless \ 2\; ,\; \; \; ODZ:\; \; \left \{ {{x-2\ \textgreater \ 0} \atop {27-x\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textgreater \ 2} \atop {x\ \textless \ 27}} \right. \; \to \\\\x\in (2,27)\\\\lg(x-2)(27-x)\ \textless \ lg10^2\\\\27x-x^2-54+2x\ \textless \ 100\\\\x^2-29x+154\ \textgreater \ 0\\\\D=225\; ,\; \; x_1=\frac{29-15}{2}=7,\; \; x_2=22\\\\Otvet:\; \; x_1=7,\; x_2=22.$