Bonjour;

A = (x+3)(x+4)

= x² +4x +3x +12

= x² +7x + 12

excuse-moi pour l'erreur

B = (y+3)(2y+8)

= 2y² +8y +6y +24

= 2y² +14y +24

C = (5-y)(5+y)

= 25 +5y -5y -y²

= 25 -y²

D = (3x+2)(3x-2)​

= 9x² -6x +6x -4

= 9x² -4

Réponse :

Développer puis réduire :

[tex]A = (x+3)(x+4)\\A = x^{2} +4x+3x+12\\A = x^{2} +7x+12[/tex]

[tex]B = (y+3)(2y+8)\\B = 2y^2+8y+6y+24\\B = 2y^2+14y+24[/tex]

[tex]C = (5-y)(5+y)\\C = 5^2-y^2\\C = 25 - y^2[/tex]

[tex]D = (3x+2)(3x-2)\\D = (3x)^{2} -2^2\\D = 9x^{2} -4[/tex]