Solve-the-inequalities-x-1-2-gt-1-Thank-you-

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

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

Answered by Skabetix last updated on 04/Jun/23

$$\left.{S}=\right]−\infty,−\frac{\mathrm{1}}{\mathrm{2}}\left[\cup\right]\frac{\mathrm{3}}{\mathrm{2}},+\infty\left[\right. \\ $$

Answered by aba last updated on 04/Jun/23

$$\mid\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\mid>\mathrm{1}\:\Leftrightarrow\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}>\mathrm{1}\:\vee\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}<−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Leftrightarrow\mathrm{x}>\frac{\mathrm{3}}{\mathrm{2}}\:\vee\:\mathrm{x}<−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by Subhi last updated on 04/Jun/23

$${x}−\frac{\mathrm{1}}{\mathrm{2}}>\mathrm{1}\:\:\:\:\:\Rrightarrow\:{x}>\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${x}−\frac{\mathrm{1}}{\mathrm{2}}<−\mathrm{1}\:\:\:\:\Rrightarrow\:{x}<\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${R}−\left[\frac{−\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right] \\ $$$${or} \\ $$$${x}^{\mathrm{2}} −{x}+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}>\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}−\frac{\mathrm{3}}{\mathrm{4}}>\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)>\mathrm{0}\: \\ $$$${x}>\frac{\mathrm{3}}{\mathrm{2}}\:\:\:,\:\:{x}>−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${or}\:{x}<\frac{\mathrm{3}}{\mathrm{2}}\:\:,\:{x}<\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${R}−\left[\frac{−\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right] \\ $$

Commented by Mastermind last updated on 15/Jun/23

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

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