Question

Доброго утра)
Подскажете с модулями?

Answer 1

$2.\; \; \left |\frac{2x-1}{x-1}\right |\ \textless \ 2\; \; \to \; \; \left \{ {{\frac{2x-1}{x-1}\ \textless \ 2} \atop {\frac{2x-1}{x-1}}\ \textgreater \ -2} \right. \; \left \{ {{\frac{2x-1}{x-1}-2\ \textless \ 0} \atop {\frac{2x-1}{x-1}+2\ \textgreater \ 0}} \right. \; \left \{ {{\frac{2x-1-2x+2}{x-1}\ \textless \ 0} \atop {\frac{2x-1+2x-2}{x-1}\ \textgreater \ 0}} \right. \\\\ \left \{ {{\frac{3}{x-1}\ \textless \ 0} \atop {\frac{4x-3}{x-1}\ \textgreater \ 0}} \right. \; \left \{ {{x-1\ \textless \ 0} \atop {x\ \textless \ \frac{3}{4}\; U\; x\ \textgreater \ 1}} \right. \; \to \; \\\\\frac{4x-3}{x-1}\ \textgreater \ 0\; \; ,\; \; +++(\frac{3}{4})---(1)+++$

$x\in (-\infty ,\frac{3}{4})U(1,+\infty )\\\\ \left \{ {{x\ \textless \ 1\; \; \; [\, x\in (-\infty ,1)\; ]} \atop {x\in (-\infty,\frac{3}{4})U(1,+\infty )}} \right. \; \; \Rightarrow \; \; x\in (-\infty ,\frac{3}{4})$

$4.\; \; |5x^2-2x+1|\ \textless \ 1\; \; \Rightarrow \; \; -1\ \textless \ 5x^2-2x+1\ \textless \ 1\; \; \Rightarrow \\\\ \left \{ {{5x^2-2x+1\ \textless \ 1} \atop {5x^2-2x+1\ \textgreater \ -1}} \right. \; \left \{ {{5x^2-2x\ \textless \ 0} \atop {5x^2-2x+2\ \textgreater \ 0}} \right. \\\\a)\; x(5x-2)\ \textless \ 0\; ,\; \; +++(0)---(\frac{2}{5})+++\\\\x\in (0,\frac{2}{5})\\\\b)\; 5x^2-2x+2\ \textgreater \ 0\\\\D/4=1-10=-9\ \textless \ 0\; \; \to \; \; 5x^2-2x+2\ \textgreater \ 0\; pri \; x\in (-\infty ,+\infty )\\\\\frac{2}{5}=0,4\; \; ;\; \; \left \{ {{x\in (0;\; 0,4)} \atop {x\in (-\infty ,+\infty )}} \right. \\\\Otvet:\; x\in (0;\; 0,4).$