salut
1) (x-3)(2x+9)=0
x-3=0 => x=3
2x+9=0 => x= -9/2 S={ -9/2 ; 3}
2) A= (2x-5)(x+2)+(2x-5)(4x-6)
= (2x-5)(x+2+4x-6)
= (2x-5)(5x-4)
3) B= (x-4)6x-(2x-3)(x-4)
= (x-4)(6x-2x+3)
= (x-4)(4x+3)
4) C= 4x²+12x+9 => identité remarquable (a+b)²
= (2x+3)²
5) D= x(x-3)²-(2x+7)²
= x(x²-6x+9)-(4x²+28x+49)
= x^3-6x²+9x-4x²-28x-49
= x^3-10x²-19x-49
exo2) au cas ou si tu as besoin
f(1)=17/3 f(4)=23/3
f est de la forme y=ax+b
calcul de a
a=((23/3)-(17/3))/(4-1)
a =(6/3)/3 => 2/3
calcul de b
17/3=(2/3)*1+b
(17/3)-(2/3)=b => b=5
f(x)= (2/3)x+5
