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Question Number 81139 by TawaTawa last updated on 09/Feb/20

Commented by TawaTawa last updated on 09/Feb/20

$$\mathrm{Please}\mathrm{help}\mathrm{me}.$$

Commented by mr W last updated on 09/Feb/20

$$onlyx>1iscorrect.othersarewrong.$$$$butdoesthishelpyouanyhow?$$

Commented by TawaTawa last updated on 09/Feb/20

$$\mathrm{How}\mathrm{is}\mathrm{the}\mathrm{answer}\mathrm{x}>1\mathrm{sir}?.$$

Commented by msup trace by abdo last updated on 09/Feb/20

$$ifx>0weget1<x\Rightarrow x>1$$$$ifx<0weget-1<x\Rightarrow -1⩽x<0$$

Commented by TawaTawa last updated on 09/Feb/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by mathmax by abdo last updated on 10/Feb/20

$$youarewelcome.$$

Answered by Kamel Kamel last updated on 09/Feb/20

$$\mathrm{\forall }x\in \mathbb{R},\mid x\mid ⩾0,then\frac{x}{\mid x\mid }<x\iff x<x\mid x\mid ..\left(E\right)$$$$Ifx>0,\left(E\right)\Rightarrow x<x^2\Rightarrow x>1.$$$$Ifx<0,\left(E\right)\Rightarrow x<-x^2\Rightarrow 1>-x\Rightarrow 0>x>-1(multiplingby\frac{1}{x}<0)$$$$Thenx>1istrue.$$

Commented by TawaTawa last updated on 09/Feb/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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