Chapitre 6 - Equations et In´equations

(1)

GYMNASE DE BURIER

Chapitre 6 - Equations et In´equations

Sarah D´egallier Rochat

(2)

1. Equations

Exemple 1.1 R´esoudre l’´equation15 + 3x= 1−4x 15 + 3x = 1−4x

+4x

⇔ 15 + 3x+4x

= 1−4x+4x CL^∗

⇔ 15 + 7x

= 1 −15

⇔ 15 + 7x−15

= 1−15 CL

⇔ 7x

= −14 ÷7

=·1 7

⇔ 1 7·7·x

= 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

ImportantUn seul signe ´egal par ligne !

∗ CL = calcul litt´eral

(3)

1. Equations

Exemple 1.1 R´esoudre l’´equation15 + 3x= 1−4x 15 + 3x = 1−4x +4x

⇔ 15 + 3x+4x

= 1−4x+4x CL^∗

⇔ 15 + 7x

= 1 −15

⇔ 15 + 7x−15

= 1−15 CL

⇔ 7x

= −14 ÷7

=·1 7

⇔ 1 7·7·x

= 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(4)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x

CL^∗

⇔ 15 + 7x

= 1 −15

⇔ 15 + 7x−15

= 1−15 CL

⇔ 7x

= −14 ÷7

=·1 7

⇔ 1 7·7·x

= 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(5)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x

= 1 −15

⇔ 15 + 7x−15

= 1−15 CL

⇔ 7x

= −14 ÷7

=·1 7

⇔ 1 7·7·x

= 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(6)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x

= 1 −15

⇔ 15 + 7x−15

= 1−15 CL

⇔ 7x

= −14 ÷7

=·1 7

⇔ 1 7·7·x

= 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(7)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1

−15

⇔ 15 + 7x−15

= 1−15 CL

⇔ 7x

= −14 ÷7

=·1 7

⇔ 1 7·7·x

= 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(8)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15

= 1−15 CL

⇔ 7x

= −14 ÷7

=·1 7

⇔ 1 7·7·x

= 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(9)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15

CL

⇔ 7x

= −14 ÷7

=·1 7

⇔ 1 7·7·x

= 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(10)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15 CL

⇔ 7x

= −14 ÷7

=·1 7

⇔ 1 7·7·x

= 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(11)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15 CL

⇔ 7x

= −14 ÷7

=·1 7

⇔ 1 7·7·x

= 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(12)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15 CL

⇔ 7x = −14

÷7

=·1 7

⇔ 1 7·7·x

= 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(13)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15 CL

⇔ 7x = −14 ÷7

=·1 7

⇔ 1 7·7·x

= 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(14)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15 CL

⇔ 7x = −14 ÷7=·1 7

⇔ 1 7·7·x

= 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(15)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15 CL

⇔ 7x = −14 ÷7=·1 7

⇔ 1

7·7·x = 1

7 ·(−14)

CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(16)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15 CL

⇔ 7x = −14 ÷7=·1 7

⇔ 1

7·7·x = 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(17)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15 CL

⇔ 7x = −14 ÷7=·1 7

⇔ 1

7·7·x = 1

7 ·(−14) CL

⇔ 7 7·x

= −14 7

CL

⇔ x

= −2

⇒S={−2}

(18)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15 CL

⇔ 7x = −14 ÷7=·1 7

⇔ 1

7·7·x = 1

7 ·(−14) CL

⇔ 7

7·x = −14 7

CL

⇔ x

= −2

⇒S={−2}

(19)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15 CL

⇔ 7x = −14 ÷7=·1 7

⇔ 1

7·7·x = 1

7 ·(−14) CL

⇔ 7

7·x = −14

7 CL

⇔ x

= −2

⇒S={−2}

(20)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15 CL

⇔ 7x = −14 ÷7=·1 7

⇔ 1

7·7·x = 1

7 ·(−14) CL

⇔ 7

7·x = −14

7 CL

⇔ x

= −2

⇒S={−2}

(21)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15 CL

⇔ 7x = −14 ÷7=·1 7

⇔ 1

7·7·x = 1

7 ·(−14) CL

⇔ 7

7·x = −14

7 CL

⇔ x = −2

⇒S={−2}

(22)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15 CL

⇔ 7x = −14 ÷7=·1 7

⇔ 1

7·7·x = 1

7 ·(−14) CL

⇔ 7

7·x = −14

7 CL

⇔ x = −2

⇒S={−2}

(23)

1. Equations

⇔ 15 + 3x+4x = 1−4x+4x CL^∗

⇔ 15 + 7x = 1 −15

⇔ 15 + 7x−15 = 1−15 CL

⇔ 7x = −14 ÷7=·1 7

⇔ 1

7·7·x = 1

7 ·(−14) CL

⇔ 7

7·x = −14

7 CL

⇔ x = −2

⇒S={−2}

(24)

Exemple 1.2 R´esoudre 5x+ 3 = 4 + 5x 5x+ 3 = 4 + 5x

−5x

⇔ 5x−5x+ 3

= 4 + 5x−5x CL

⇔ 0·x+ 3

= 4 + 0·x CL

⇔ 3

= 4

Cette ´equation n’a pas de solution !

On noteS=∅

Exemple 1.3 R´esoudre x−1 =x−1

x−1 = x−1

−x

⇔ x−x−1

= x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

Cette ´equation est vraie quel que soitx!

On noteS=R

(25)

Exemple 1.2 R´esoudre 5x+ 3 = 4 + 5x

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3

= 4 + 5x−5x CL

⇔ 0·x+ 3

= 4 + 0·x CL

⇔ 3

= 4

On noteS=∅

x−1 = x−1

−x

⇔ x−x−1

= x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On noteS=R

(26)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x

CL

⇔ 0·x+ 3

= 4 + 0·x CL

⇔ 3

= 4

On noteS=∅

x−1 = x−1

−x

⇔ x−x−1

= x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On noteS=R

(27)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3

= 4 + 0·x CL

⇔ 3

= 4

On noteS=∅

x−1 = x−1

−x

⇔ x−x−1

= x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On noteS=R

(28)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3

= 4 + 0·x CL

⇔ 3

= 4

On noteS=∅

x−1 = x−1

−x

⇔ x−x−1

= x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On noteS=R

(29)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x

CL

⇔ 3

= 4

On noteS=∅

x−1 = x−1

−x

⇔ x−x−1

= x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On noteS=R

(30)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3

= 4

On noteS=∅

x−1 = x−1

−x

⇔ x−x−1

= x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On noteS=R

(31)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4

On noteS=∅

x−1 = x−1

−x

⇔ x−x−1

= x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On noteS=R

(32)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4 Cette ´equation n’a pas de solution !

On note S=∅ Exemple 1.3 R´esoudre x−1 =x−1

x−1 = x−1

−x

⇔ x−x−1

= x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On noteS=R

(33)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4

Cette ´equation n’a pas de solution ! On note S=∅

x−1 = x−1

−x

⇔ x−x−1

= x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On noteS=R

(34)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4

Exemple 1.3 R´esoudre x−1 =x−1 x−1 = x−1

−x

⇔ x−x−1

= x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On note S=R

(35)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4

x−1 = x−1 −x

⇔ x−x−1

= x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On note S=R

(36)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4

x−1 = x−1 −x

⇔ x−x−1

= x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On note S=R

(37)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4

x−1 = x−1 −x

⇔ x−x−1 = x−x−1

CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On note S=R

(38)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4

x−1 = x−1 −x

⇔ x−x−1 = x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On note S=R

(39)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4

x−1 = x−1 −x

⇔ x−x−1 = x−x−1 CL

⇔ 0·x−1

= 0·x−1 CL

⇔ −1

= −1

On note S=R

(40)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4

x−1 = x−1 −x

⇔ x−x−1 = x−x−1 CL

⇔ 0·x−1 = 0·x−1

CL

⇔ −1

= −1

On note S=R

(41)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4

x−1 = x−1 −x

⇔ x−x−1 = x−x−1 CL

⇔ 0·x−1 = 0·x−1 CL

⇔ −1

= −1

On note S=R

(42)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4

x−1 = x−1 −x

⇔ x−x−1 = x−x−1 CL

⇔ 0·x−1 = 0·x−1 CL

⇔ −1

= −1 Cette ´equation est vraie quel que soitx!

On note S=R

(43)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4

x−1 = x−1 −x

⇔ x−x−1 = x−x−1 CL

⇔ 0·x−1 = 0·x−1 CL

⇔ −1 = −1

On note S=R

(44)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4

x−1 = x−1 −x

⇔ x−x−1 = x−x−1 CL

⇔ 0·x−1 = 0·x−1 CL

⇔ −1 = −1 Cette ´equation est vraie quel que soitx!

On note S=R

(45)

5x+ 3 = 4 + 5x −5x

⇔ 5x−5x+ 3 = 4 + 5x−5x CL

⇔ 0·x+ 3 = 4 + 0·x CL

⇔ 3 = 4

x−1 = x−1 −x

⇔ x−x−1 = x−x−1 CL

⇔ 0·x−1 = 0·x−1 CL

⇔ −1 = −1

Cette ´equation est vraie quel que soitx! On note S=R