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
![solution: this is equivalent to ∣((x^2 +1)/x)∣ > 2 ((x^2 +1)/(∣x∣)) > 2 [ since x^2 +1 >0 ] x^2 +1 > 2∣x∣ ; x^2 −2∣x∣+1 > 0 ∣x∣^2 −2∣x∣+1 > 0 , (∣x∣−1)^2 > 0 ∣x∣ ≠ 1 . answer all x except x =1 and x=−1](https://www.tinkutara.com/question/Q95637.png)
$$\mathrm{solution}:\: \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}\:\mid\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}\mid\:>\:\mathrm{2} \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\mid{x}\mid}\:>\:\mathrm{2}\:\left[\:\mathrm{since}\:{x}^{\mathrm{2}} +\mathrm{1}\:>\mathrm{0}\:\right]\: \\ $$$${x}^{\mathrm{2}} +\mathrm{1}\:>\:\mathrm{2}\mid{x}\mid\:;\:{x}^{\mathrm{2}} −\mathrm{2}\mid{x}\mid+\mathrm{1}\:>\:\mathrm{0} \\ $$$$\mid{x}\mid^{\mathrm{2}} −\mathrm{2}\mid{x}\mid+\mathrm{1}\:>\:\mathrm{0}\:,\:\left(\mid{x}\mid−\mathrm{1}\right)^{\mathrm{2}} \:>\:\mathrm{0} \\ $$$$\mid{x}\mid\:\neq\:\mathrm{1}\:.\:\mathrm{answer}\:\mathrm{all}\:{x}\:{except}\:{x}\:=\mathrm{1}\:{and}\:{x}=−\mathrm{1} \\ $$
Commented by peter frank last updated on 26/May/20
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by i jagooll last updated on 26/May/20
$$\mathrm{thank}\:\mathrm{you} \\ $$