MATHS 1ES
1724 mots
7 pages
CHAPITRE1
Second degré
Entraînement (page 32)
EXERCICES
Ά
DE TÊTE
(2x – 3) (8 + 2x) = 0 ; = – 4 ;
50 2[(x – 3)2 – 4] = 0.
39 1. A(1) ≠ 0.
Donc 1 n’est pas solution de l’équation A(x) = 0.
2. A(x) > 0.
c) A(12) > 0.
ÉQUATIONS DU SECOND DEGRÉ
PARTICULIÈRES
Ά
·
1
(x + 3) (2x + 1) = 0 ; = – 3 ; –
.
2
42 [(x – 1) + (2 – 3x)] [(x – 1) – (2 – 3x)] = 0.
(– 2x + 1) (4x – 3) = 0 ; =
2[(x – 3 – 2) (x – 3 + 2)] = 0.
2(x – 5) (x – 1) = 0 ; = {1 ; 5}.
51 a) [3 – 2(x – 2)] [3 + 2(x – 2)] = 0.
40 1. (x – 1) (x + 2) = A(x).
2. = {– 2 ; 1}.
3. a) A(0) < 0.
b) A(– 3) > 0.
41
·
3
.
2
Ά 1 ; 3 ·.
2 4
43 (x – 4) [(x – 4) + (x + 2)] = 0.
(x – 4) (2x – 2) = 0 ; = {1 ; 4}.
(– 2x + 7) (2x – 1) = 0 ; =
b) = Ø.
Ά 1 ; 7 ·.
2 2
52 a) [512 – (2x + 3)] [512 + (2x + 3)] = 0.
(– 2x + 512 – 3) (2x + 512 + 3) = 0 ;
512 – 3 – 512 – 3
=
;
.
2
2
b) = Ø.
Ά
·
DEUX MÉTHODES DE RÉSOLUTION
53 • [2x + (x – 2)] [2x – (x – 2)] = 0.
Ά
(3x – 2) (x + 2) = 0 ; = – 2 ; –
2
2
·
2
.
3
(x + 2) [(x – 2) – 3(x + 2)] = 0.
(x + 2) (– 2x – 8) = 0 ; = {– 4 ; – 2}.
• 4x – (x – 4x + 4) = 0.
3x2 + 4x – 4 = 0.
Δ = 64, deux solutions :
–4 – 8
–4 + 8 2
= – 2 et x2 =
= . x1 =
6
6
3
2
= – 2; –
.
3
54 • (x + 3) (1 – x) = 0 ; = {– 3 ; 1}.
• x + 3 – x2 – 3x = 0.
– x2 – 2x + 3 = 0.
Δ = 16, deux solutions :
2–4
2+4 x1 =
= 1 et x2 =
= – 3.
–2
–2
= {– 3 ; 1}.
48 3(x2 – 4x + 4) = 0.
55 • 2(5 + 2 – 3x) (5 – 2 + 3x) = 0.
44 2(9x2 + 6x + 1) = 0.
Ά 1 ·.
3
2(3x + 1)2 = 0 ; = –
45 (2 + 5x) [(3x) – (x + 1)] = 0.
Ά
·
2 1
;
.
5 2
46 2x(x + 3) – (x + 3) (x – 3) = 0.
(x + 3) [2x – (x – 3)] = 0.
(x + 3) (x + 3) = 0 ; = {– 3}.
(2 + 5x) (2x – 1) = 0 ; = –
47 (x + 2) (x – 2) – 3(x + 2)2 = 0.
3(x – 2)2 = 0 ; = {2}.
49 5(2x – 3) – (3 + 2x) (3 – 2x) = 0.
Ά
·
Ά
2(7 – 3x) (3 + 3x) = 0 ; = – 1 ;
• 50 – 2(4