[tex]\dfrac{|x-1|}{x}\geq1\qquad(x\not=0)[/tex]
Ostatecznie zatem [tex]x\in\left(0,\dfrac{1}{2}\right\rangle[/tex].
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[tex]\dfrac{|x-1|}{x}\geq1\qquad(x\not=0)[/tex]
[tex]\dfrac{-x+1}{x}\geq1\\-x^2+x\geq x^2\\2x^2-x\leq0\\x(2x-1)\leq0\\x\in\left\langle0,\dfrac{1}{2}\right\rangle\\\\\\x\in\left\langle0,\dfrac{1}{2}\right\rangle \wedge x\not = 0\wedge x\in(-\infty,1\rangle\\x\in\left(0,\dfrac{1}{2}\right\rangle[/tex]
[tex]\dfrac{x-1}{x}\geq1\\x^2-x\geq x^2\\x\leq0\\\\x\leq 0 \wedge x\not= 0 \wedge x\in(1,\infty)\\x\in\emptyset[/tex]
Ostatecznie zatem [tex]x\in\left(0,\dfrac{1}{2}\right\rangle[/tex].