Réponse :
[tex]3(x+4)+2 = 6\\3x + 12 + 2 = 6\\3x + 12 = 4\\3x = -8\\\frac{3x}{3} = \frac{-8}{3}\\\\x = -2,66[/tex]
[tex]5(x-3) - 2x(x-3) = 0\\5x - 15 - 2x^2 - 6x = 0\\5x - 2x^2 - 6x = 15\\-1x - 2x^2 = 15\\-1x - 2x = 3,87\\\frac{-3x}{-3} = \frac{3.87}{-3}\\\\x = -1.29[/tex]
[tex](2x - 2)^2 = 4(x^2-2x)\\2x - 2*2x*2 - 2^2 = 4x^2 - 8x\\2x - 8x - 4 = 4x^2 - 8x\\2x - 4 = 4x^2\\-4 = 4x^2 - 2x\\\frac{2i}{2} = \frac{2x}{2}\\\\x= i[/tex]
Problème : x ∈ [tex]\begin{document}\begin{tabular}{ll}$\mathbb{R}$\end{tabular}\end{document}[/tex]
