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Question Number 171025 by Mastermind last updated on 06/Jun/22

Answered by greougoury555 last updated on 06/Jun/22

$$\left(1\right)x⩾-3\Rightarrow \frac{2x+3-x-2}{x+2}>0$$$$\Rightarrow \frac{x+1}{x+2}>0;x<-2\cup x>-1$$$$\therefore solution:[-3,-2)\cup (-1,\mathrm{\infty })$$$$\left(2\right)x<-3\Rightarrow \frac{-3-x-2}{x+2}>0$$$$\Rightarrow \frac{-x-5}{x+2}>0;\frac{x+5}{x+2}<0$$$$\Rightarrow -5<x<-2$$$$\therefore solution:(-5,-3)$$$$\iff \therefore finalsolution\equiv (-5,-2)\cup (-1,\mathrm{\infty })$$

Commented by Tawa11 last updated on 07/Jun/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by thfchristopher last updated on 07/Jun/22

$$\Rightarrow \left(\frac{\mid x+3\mid +x}{x+2}\right)\left(\frac{x+2}{x+2}\right)>1$$$$\Rightarrow (x+2)\mid x+3\mid +x^2+2x>x^2+4x+4$$$$\Rightarrow (x+2)\mid x+3\mid -2(x+2)>0$$$$\Rightarrow (x+2)(\mid x+3\mid -2)>0$$$$\mathrm{Case}1:$$$$\mid x+3\mid -2>0\mathrm{and}x+2>0\Rightarrow x>-2$$$$\Rightarrow \mid x+3\mid >2$$$$\Rightarrow x+3<-2\mathrm{or}x+3>2$$$$\Rightarrow x<-5\left(\mathrm{rejected}\right)\mathrm{or}x>-1$$$$\therefore x>-1$$$$\mathrm{Case}2:$$$$\mid x+3\mid -2<0\mathrm{and}x+2<0\Rightarrow x<-2$$$$\Rightarrow \mid x+3\mid <2$$$$\Rightarrow -2<x+3<2$$$$\Rightarrow -5<x<-1$$$$\therefore -5<x<-2$$$$\mathrm{Conclusion}:-5<x<-2\mathrm{or}x>-1$$

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