Bonjour,
1)
A =25x²+16-40x-(x-3)(5x-4)
A = 25x²+16-40x-(5x²-4x-15x+20)=
A = 25x²+16-40x-5x²+4x+15x-20=
A = 20x²-21x-4
2a)
C'est une identité remarquable de la forme a²+2ab+b² = (a+b)²
25x²+16-40x = (5x-4)²
2b)
A = 25x²+16-40x-(x-3)(5x-4)
A = (5x-4)²-(x-3)(5x-4)
A = (5x-4)(5x-4-(x-3))
A = (5x-4)(5x-4-x+3)
A = (5x-4)(4x-1)
3)
A = (5x-4)(4x-1) --> Pour que A = 0 il suffit qu'un des facteurs soit nul :
5x-4 = 0 --> x = 4/5
4x-1 = 0 --> x = 1/4
S = {1/4 ; 4/5}
4)
si x = 9/5
A = (5x-4)(4x-1) =
A = (5*9/5-4)(4*9/5-1)
A = (9-4)(36/5-5/5)=
A = 5(31/5)
A = 31
J'espère que tu as compris
a+
