Ответ(ы) на вопрос:
Гость [latex]|x|^{x^2-3x+2}\ \textless \ 1[/latex] ,
Прологарифмируем неравенство по любому основанию.
[latex]lg(|x|^{x^2-3x+2})\ \textless \ lg1\; ,\; \; \; |x| \geq 0\\\\(x^2-3x+2)\cdot lg|x|\ \textless \ 0\\\\x^2-3x+2=0\; \; \Rightarrow \; \; \; x_1=1,\; \; x_2=2\; \; (teor.\; Vieta)\\\\x^2-3x+2=(x-1)(x-2)\\\\1)\; \; \left \{ {{(x-1)(x-2)\ \textless \ 0} \atop {lg|x|\ \textgreater \ 0}} \right. \; ,\; \left \{ {{1\ \textless \ x\ \textless \ 2} \atop {|x|\ \textgreater \ 1}} \right. \; ,\; \left \{ {{1\ \textless \ x\ \textless \ 2} \atop {x\ \textless \ -1;\; \; x\ \textgreater \ 1}} \right. \; \; \to \; \; 1\ \textless \ x\ \textless \ 2[/latex]
[latex]2)\; \; \left \{ {{(x-1)(x-2)\ \textgreater \ 0} \atop {lg|x|\ \textless \ 0}} \right. \; ,\; \left \{ {{x\ \textless \ 1;\; x\ \textgreater \ 2} \atop {|x|\ \textless \ 1}} \right. \; ,[/latex]
[latex] \left \{ {{x\ \textless \ 1;\; x\ \textgreater \ 2} \atop {-1\ \textless \ x\ \textless \ 1}} \right. \; \; \to \; \; x\in \varnothing \\\\Otvet:\; \; x\in (1,2)\; .[/latex]
2 способ. [latex]|x|^{x^2-3x+2}\ \textless \ |x|^0[/latex] . [latex]|x|^0=1[/latex]
[latex]1)|x|\ \textgreater \ 1\to \left [ {{x\ \textgreater \ 1} \atop {x\ \textless \ -1}} \right. ;\\\\x^2-3x+2\ \textless \ 0;(x-1)(x-2)\ \textless \ 0;\\\\1\ \textless \ x\ \textless \ 2\\\\2)|x|\ \textless \ 1;-1\ \textless \ x\ \textless \ 1\\\\x^2-3x+2\ \textgreater \ 0;(x-1)(x-2)\ \textgreater \ 0; \left [ {{-1\ \textless \ x\ \textless \ 1} \atop {x\ \textless \ 1,x\ \textgreater \ 2}} \right. \to x\in \varnothing \\\\3)|x|=1\to x=\pm 1[/latex]
При подстановке х= -1 или х=1 не получаем верного равенства.
Ответ: 1
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