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Question Number 95634 by i jagooll last updated on 26/May/20 | ||
$$\mathrm{solve}\:\mid{x}+\frac{\mathrm{1}}{{x}}\mid\:>\:\mathrm{2}\: \\ $$ | ||
Answered by john santu last updated on 26/May/20 | ||
$$\mathrm{solution}:\: \\ $$ $$\mathrm{this}\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}\:\mid\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}\mid\:>\:\mathrm{2} \\ $$ $$\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\mid{x}\mid}\:>\:\mathrm{2}\:\left[\:\mathrm{since}\:{x}^{\mathrm{2}} +\mathrm{1}\:>\mathrm{0}\:\right]\: \\ $$ $${x}^{\mathrm{2}} +\mathrm{1}\:>\:\mathrm{2}\mid{x}\mid\:;\:{x}^{\mathrm{2}} −\mathrm{2}\mid{x}\mid+\mathrm{1}\:>\:\mathrm{0} \\ $$ $$\mid{x}\mid^{\mathrm{2}} −\mathrm{2}\mid{x}\mid+\mathrm{1}\:>\:\mathrm{0}\:,\:\left(\mid{x}\mid−\mathrm{1}\right)^{\mathrm{2}} \:>\:\mathrm{0} \\ $$ $$\mid{x}\mid\:\neq\:\mathrm{1}\:.\:\mathrm{answer}\:\mathrm{all}\:{x}\:{except}\:{x}\:=\mathrm{1}\:{and}\:{x}=−\mathrm{1} \\ $$ | ||
Commented bypeter frank last updated on 26/May/20 | ||
$$\mathrm{thank}\:\mathrm{you} \\ $$ | ||
Commented byi jagooll last updated on 26/May/20 | ||
$$\mathrm{thank}\:\mathrm{you} \\ $$ | ||