Bonjour vous pouvez m'aider c'est un dm de maths svp ?

Salut,

Alors pour la première :

[tex](1+4x)^{2} = (8x +1)(2x - 3) \\ 1+8x + 16x^{2} = 16 x^{2} - 24x +2x -3 \\ 1+3+16 x^{2} - 16 x^{2} +22x +8x = 0 \\ 4 + 30x = 0 \\ 30x = -4 \\ x = \frac{-2}{15} [/tex]

La deuxième : [tex] \frac{3-7x}{9} - \frac{2x - 1}{3} = \frac{5x + 2}{18} \\ \frac{6-14x - (12x -6) - (5x + 2)}{18} = 0 \\ \frac{6-14x - 12x + 6 -5x -2 }{18} = 0 \\ \frac{-31x+10}{18} =0 \\ -31x + 10 = 0 \\ x = \frac{10}{31} [/tex]

[tex](2x + 3)(5-3x) = 4x^{2} -9 \\ (2x + 3)(5-3x) = (2x+3)(2x - 3) \\ (2x + 3)(5-3x) - (2x + 3)(2x - 3) = 0 \\ (2x +3) [5-3x - (2x -3)] = 0 \\ (2x + 3)(5-3x - 2x +3) = 0 \\ (2x + 3) (8-5x) = 0 \\ 2x + 3 = 0 \\ 2x = -3 \\ x = \frac{-3}{2} \\ ou \\ 8-5x = 0 \\ 8 = 5x \\ x = \frac{8}{5} [/tex]

[tex](3-5x)^{2} = (6-x)^{2} \\ (3-5x)^{2} - (6-x)^{2} = 0 \\ (3-5x + 6 - x)(3-5x - 6 +x) = 0 \\ (9 - 6x)(-3 -4x) = 0 \\ 9-6x = 0 \\ 9 = 6x \\ x = \frac{9}{6} \\ x = \frac{3}{2} \\ OU \\ -3 - 4x = 0 \\ -3 = 4x \\ x = -\frac{3}{4} [/tex]