Réponses

2014-02-14T21:36:50+01:00
Bonsoir,

a) 12x^2-18=0\\6(2x^2-3)=0\\2x^2-3=0\\(\sqrt{2}x-\sqrt{3})(\sqrt{2}x+\sqrt{3})=0\\\\\sqrt{2}x-\sqrt{3}=0\ \ ou\ \ \sqrt{2}x+\sqrt{3}=0\\\\\sqrt{2}x=\sqrt{3}\ \ ou\ \ \sqrt{2}x=-\sqrt{3}\\\\x=\dfrac{\sqrt{3}}{\sqrt{2}}\ \ ou\ \ x=-\dfrac{\sqrt{3}}{\sqrt{2}}\\\\x=\dfrac{\sqrt{3}\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}\ \ ou\ \ x=-\dfrac{\sqrt{3}\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}\\\\x=\dfrac{\sqrt{6}}{2}\ \ ou\ \ x=-\dfrac{\sqrt{6}}{2}


b) 3x (2x - 1)(-5x + 7)  = 0
3x = 0   ou   2x - 1 = 0   ou   -5x + 7 = 0
x = 0   ou   2x = 1   ou   -5x = -7
x = 0   ou   x = 1/2   ou   x = (-7)/(-5)
x = 0   ou   x = 1/2   ou   x = 7/5