quelqu'un saurait faire: [tex] x \sqrt{2}-7=5x+6[/tex] [tex]3(2x-1) \leq 4(3x-3)+15[/tex] [tex] \frac{8-x}{6}+ \frac{2(x-1)}{3} < \frac{x+6}{2}- \frac{x}{3}

xV2 - 5x = +6 + 7 => (V2 - 5)x = 13 => x = 13/(V2 - 5) = 13(V2 + 5)/-23
6x - 3 - 12x + 12 - 15 <=0 => -6x -6 <= 0 => -6x <= 6 => x >= - 1
8 - x + 4x - 4 < 3x + 18 - 2x => 3x + 4 < x + 18  => 2x < 14 => x < 7