Bonjour Sisilastar74
[tex]-3(2x+5)-3x+1\ \textgreater \ x(-5x+5)\\\\-6x-15-3x+1\ \textgreater \ -5x^2+5x\\\\-9x-15+1\ \textgreater \ -5x^2+5x\\\\5x^2-5x-9x-15+1\ \textgreater \ 0\\\\5x^2-14x-14\ \textgreater \ 0[/tex]
Tableau de signes de 5x² - 14x - 14.
Racines :
[tex]5x^2 - 14x - 14 = 0\\\\5(x^2-\dfrac{14}{5}x-\dfrac{14}{5})=0\\\\x^2-\dfrac{14}{5}x-\dfrac{14}{5}=0\\\\x^2-2\times\dfrac{7}{5}x-\dfrac{14}{5}=0\\\\ \ [x^2-2\times\dfrac{7}{5}x+(\dfrac{7}{5})^2]-(\dfrac{7}{5})^2-\dfrac{14}{5}=0\\\\(x-\dfrac{7}{5})^2-(\dfrac{7}{5})^2-\dfrac{14}{5}=0\\\\(x-\dfrac{7}{5})^2-\dfrac{49}{25}-\dfrac{14}{5}=0\\\\(x-\dfrac{7}{5})^2-\dfrac{119}{25}=0[/tex]
[tex](x-\dfrac{7}{5})^2-(\dfrac{\sqrt{119}}{5})^2=0\\\\(x-\dfrac{7}{5}+\dfrac{\sqrt{119}}{5})(x-\dfrac{7}{5}-\dfrac{\sqrt{119}}{5})=0\\\\(x-\dfrac{7-\sqrt{119}}{5})(x-\dfrac{7+\sqrt{119}}{5})=0\\\\x-\dfrac{7-\sqrt{119}}{5}=0\ \ ou\ \ x-\dfrac{7+\sqrt{119}}{5}=0\\\\x=\dfrac{7-\sqrt{119}}{5}\ \ ou\ \ x=\dfrac{7+\sqrt{119}}{5}[/tex]
De plus, nous avons : [tex]5x^2-14x-14=5(x-\dfrac{7-\sqrt{119}}{5})(x-\dfrac{7+\sqrt{119}}{5})[/tex]
D'où le tableau :
[tex]\begin{array}{|c|ccccccc|} x&-\infty&&\dfrac{7-\sqrt{119}}{5}&&\dfrac{7+\sqrt{119}}{5}&&+\infty \\ 5&&+&+&+&+&+&\\x-\dfrac{7-\sqrt{119}}{5}&&-&0&+&+&+&\\x-\dfrac{7+\sqrt{119}}{5}&&-&-&-&0&+&\\5x^2-14x-14&&+&0&-&0&+&\\ \end{array}\\\\\\5x^2 - 14x - 14\ \textgreater \ 0\Longleftrightarrowx\in\ ]-\infty;\dfrac{7-\sqrt{119}}{5}[\ \cup\ ]\dfrac{7+\sqrt{119}}{5};+\infty[[/tex]
Par conséquent,
l'ensemble des solutions de l'inéquation est [tex]\boxed{S= ]-\infty;\dfrac{7-\sqrt{119}}{5}[\ \cup\ ]\dfrac{7+\sqrt{119}}{5};+\infty[}[/tex]
