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Question Number 27976 by NECx last updated on 18/Jan/18

solve    ((2x)/(x^2 +1))<((3x+1)/(2(x^2 +1)))

$${solve} \\ $$ $$ \\ $$ $$\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}<\frac{\mathrm{3}{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$ $$ \\ $$

Commented byabdo imad last updated on 18/Jan/18

due to   x^2 +1>0   ( e)  ⇔   2x <   ((3x+1)/2)  ⇔  4x −3x−1<0 ⇔  x−1<0  ⇔  x <1   ⇔ x∈ ]−∝ ,1[

$${due}\:{to}\:\:\:{x}^{\mathrm{2}} +\mathrm{1}>\mathrm{0}\:\:\:\left(\:{e}\right)\:\:\Leftrightarrow\:\:\:\mathrm{2}{x}\:<\:\:\:\frac{\mathrm{3}{x}+\mathrm{1}}{\mathrm{2}} \\ $$ $$\left.\Leftrightarrow\:\:\mathrm{4}{x}\:−\mathrm{3}{x}−\mathrm{1}<\mathrm{0}\:\Leftrightarrow\:\:{x}−\mathrm{1}<\mathrm{0}\:\:\Leftrightarrow\:\:{x}\:<\mathrm{1}\:\:\:\Leftrightarrow\:{x}\in\:\right]−\propto\:,\mathrm{1}\left[\right. \\ $$

Answered by Rasheed.Sindhi last updated on 18/Jan/18

x^2 +1>0  Multiplying by 2(x^2 +1) to both  sides  ((2x)/(x^2 +1))<((3x+1)/(2(x^2 +1)))⇒4x<3x+1  x<1

$$\mathrm{x}^{\mathrm{2}} +\mathrm{1}>\mathrm{0} \\ $$ $$\mathrm{Multiplying}\:\mathrm{by}\:\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\:\mathrm{to}\:\mathrm{both} \\ $$ $$\mathrm{sides} \\ $$ $$\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}<\frac{\mathrm{3}{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\Rightarrow\mathrm{4x}<\mathrm{3x}+\mathrm{1} \\ $$ $$\mathrm{x}<\mathrm{1} \\ $$

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