bonjour
1)algorithme
a)
il faut remplacer x = 1/3
(2x-1)(2x-3)
=(2× 1/3 - 1 ) ( 2× 1/3 -3)
=(2/3 -3/3)(2/3 -9/3) = -1/3 × -7/3
= 7/9
il faut remplacer x = √2
(2x-1)(2x-3)
=(2×√2 -1 ) ( 2×√2 -3)
=2√2 × 2√2 - 6√2 - 2√2 + 3 = 8 - 8√2+3
= 11 -8√2
2)
a)
f(x) =(2x-1)(2x-3)
b)
(2x-1)(2x-3) = 4x²-6x-2x+3
=4x²-8x+3
c)
on développe 4(x-1)² -1
= 4(x²-2x+1) -1
=4x²-8x+4-1
=4x²-8x+3 = f(x)
3)
a)
f(0) = 4×0² -8×0 +3
= 3
b)
f(x) = 0 =>
f(x) =(2x-1)(2x-3)
2x-1 = 0 =>
x = 1/2
ou
2x-3 =0 =>
3/2
solutions ={ ½ ; 3/2}
c)
f(x) = 3
4x²-8x+3 = 3
4x²-8x = 0
=> x(4x-8) = 0 =>
x = 0
ou 4x -8 =0
=> 4x = 8 => x = 8/4 => x =2
les antécédents de 3 sont 0 et 2
d)
on utilise la
forme canonique 4(x-1)² -1
minimum = - 1
il est atteint pour x = 1
