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Question Number 193139 by Mastermind last updated on 04/Jun/23

$$\mathrm{Solve}\mathrm{the}\mathrm{inequalities}:$$$$\mid \mathrm{x}-\frac{1}{2}\mid >1$$$$$$$$\mathrm{Thank}\mathrm{you}$$

Answered by Skabetix last updated on 04/Jun/23

$$S=]-\mathrm{\infty },-\frac{1}{2}[\cup ]\frac{3}{2},+\mathrm{\infty }[$$

Answered by aba last updated on 04/Jun/23

$$\mid \mathrm{x}-\frac{1}{2}\mid >1\iff \mathrm{x}-\frac{1}{2}>1\vee \mathrm{x}-\frac{1}{2}<-1$$$$\iff \mathrm{x}>\frac{3}{2}\vee \mathrm{x}<-\frac{1}{2}$$

Answered by Subhi last updated on 04/Jun/23

$$x-\frac{1}{2}>1\Rrightarrow x>\frac{3}{2}$$$$x-\frac{1}{2}<-1\Rrightarrow x<\frac{-1}{2}$$$$R-[\frac{-1}{2},\frac{3}{2}]$$$$or$$$$x^2-x+\frac{1}{4}-1>0$$$$x^2-x-\frac{3}{4}>0$$$$(x-\frac{3}{2})(x+\frac{1}{2})>0$$$$x>\frac{3}{2},x>-\frac{1}{2}$$$$orx<\frac{3}{2},x<\frac{-1}{2}$$$$R-[\frac{-1}{2},\frac{3}{2}]$$

Commented by Mastermind last updated on 15/Jun/23

$$\mathrm{Thank}\mathrm{you}!$$

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