Bonsoir,
1) [tex]-2x+\dfrac{1}{3}>0\Longleftrightarrow -2x>-\dfrac{1}{3}\Longleftrightarrow x <\dfrac{-\dfrac{1}{3}}{-2}\Longleftrightarrow x <\dfrac{1}{6}[/tex]
[tex]-2x+\dfrac{1}{3}=0\Longleftrightarrow -2x=-\dfrac{1}{3}\Longleftrightarrow x =\dfrac{-\dfrac{1}{3}}{-2}\Longleftrightarrow x =\dfrac{1}{6}[/tex]
[tex]-2x+\dfrac{1}{3}<0\Longleftrightarrow -2x<-\dfrac{1}{3}\Longleftrightarrow x >\dfrac{-\dfrac{1}{3}}{-2}\Longleftrightarrow x >\dfrac{1}{6}[/tex]
Donc,
[tex]Si\ \ x\in]-\infty;\dfrac{1}{6}[,\ \ alors\ \ -2x+\dfrac{1}{3}>0\\\\Si\ \ x=\dfrac{1}{6},\ \ alors\ \ -2x+\dfrac{1}{3}=0\\\\Si\ \ x\in]\dfrac{1}{6};+\infty[,\ \ alors\ \ -2x+\dfrac{1}{3}<0[/tex]
2) [tex]1+\dfrac{x}{3}<0\Longleftrightarrow \dfrac{x}{3}<-1\Longleftrightarrow x<-3\\\\1+\dfrac{x}{3}=0\Longleftrightarrow \dfrac{x}{3}=-1\Longleftrightarrow x=-3\\\\1+\dfrac{x}{3}>0\Longleftrightarrow \dfrac{x}{3}>-1\Longleftrightarrow x>-3[/tex]
Donc,
[tex]Si\ x\in]-\infty;-3[,\ \ alors\ \ 1+\dfrac{x}{3}<0\\\\Si\ x=-3,\ \ alors\ \ 1+\dfrac{x}{3}=0\\\\Si\ x\in]-3;+\infty[,\ \ alors\ \ 1+\dfrac{x}{3}>0[/tex]
