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
$${only}\:{x}>\mathrm{1}\:{is}\:{correct}.\:{others}\:{are}\:{wrong}. \\ $$$${but}\:{does}\:{this}\:{help}\:{you}\:{any}\:{how}? \\ $$
Commented by TawaTawa last updated on 09/Feb/20
$$\mathrm{How}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:\:\:\mathrm{x}\:>\:\mathrm{1}\:\:\mathrm{sir}?. \\ $$
Commented by msup trace by abdo last updated on 09/Feb/20
$${if}\:{x}>\mathrm{0}\:\:{we}\:{get}\:\:\mathrm{1}<{x}\:\Rightarrow{x}>\mathrm{1} \\ $$$${if}\:{x}<\mathrm{0}\:{we}\:{get}\:−\mathrm{1}<{x}\:\Rightarrow−\mathrm{1}\leqslant{x}<\mathrm{0} \\ $$
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
$${you}\:{are}\:{welcome}. \\ $$
Answered by Kamel Kamel last updated on 09/Feb/20
$$\forall{x}\in\mathbb{R},\:\mid{x}\mid\geqslant\mathrm{0},\:{then}\:\frac{{x}}{\mid{x}\mid}<{x}\Leftrightarrow{x}<{x}\mid{x}\mid..\left({E}\right) \\ $$$${If}\:{x}>\mathrm{0},\:\left({E}\right)\Rightarrow{x}<{x}^{\mathrm{2}} \Rightarrow{x}>\mathrm{1}. \\ $$$${If}\:{x}<\mathrm{0},\:\left({E}\right)\Rightarrow{x}<−{x}^{\mathrm{2}} \Rightarrow\mathrm{1}>−{x}\Rightarrow\mathrm{0}>{x}>−\mathrm{1}\:\left({multipling}\:{by}\:\frac{\mathrm{1}}{{x}}<\mathrm{0}\right) \\ $$$${Then}\:{x}>\mathrm{1}\:{is}\:{true}. \\ $$
Commented by TawaTawa last updated on 09/Feb/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$