La valeur absolue d’un nombre est sa distance a partirde l’origin.

| 6 | = 6

Résoudre: | x - 2 | = 6 donc| x - 2 | = 6 d’ou x - 2 = 6 ou -(x - 2) = 6

x - 2 = 6 x = 8

-(x - 2) = 6 -x + 2 = 6 -x = 4 x = -4

| 8 - 2 | = 6 | 6 | = 6 6 = 6

| - 4 - 2 | = 6 | - 6 | = 6 6 = 6

Vérification: Équations valeur absolue doitÊtre vérifié.

| -6 | = 6

Les solution sont x = 8 et x = -4.

Equations

Résous | x - 3 | - | 3x + 7 | = 0

| x - 3 | = | 3x + 7 |

x - 3 = 3x + 7 -2x = 10 x = - 5

x - 3 = -(3x + 7)x - 3 = -3x - 7 4x = -4 x = -1

Vérifier:

| -5 - 3 | - | 3(-5) + 7 | = 0 | -8 | - | 8 | = 0 8 - 8 = 0 0 = 0

| -1 - 3 | - | 3(-1) + 7 | = 0 | -4 | - | 4| = 0 4 - 4 = 0 0 = 0

Alors, les solutions sontx = -5 et x = -1.

Il y a 4 cas possibles :

-(x - 3) = -(3x + 7) -x + 3 = -3x - 7 2x = -10 x = -5

-(x - 3) = 3x + 7 -x + 3 = 3x + 7 -4x = 4 x = -1

+ +, - -, + -, - +

Deux donnenet le mêmerésultat.

Solving an Absolute Value Equation

| x - 3 | - | 3x + 7 | = 0

alors,x = -1 et x = -5.

y = | x

- 3 |

- |3 x

+ 7

|

- 5 - 1

Solution par Graphe

a) | x - 2 | = 6

La solution de L’équation est l’intersectionDes deux graphes:y = | x - 2 | ety = 6.

alorsx = -4 et x = 8.

-4 8y =

| x -

2 |

y = 6

Graphe

La solution estIntersections avec l’axe des x.

Alors, les solutionssont x = -5 et x = -1.

b) | x - 3 | - | 3x + 7 | = 0

c) | x - 2 | = 2x - 1

La solution est x = 1.

y = | x

- 2 |

y =

2 x

-1

d) | x - 1 | + | x - 4 | = 7La solution est l’intersection de:y = | x - 1 | + | x - 4 | et y = 7

Les solution sont x = -1 et x = 6.

y = 7

y =

| x -

1 | +

| x

- 4 |

| x - 5 | = 2

+ (x - 5) = 2 x - 5 = 2 x = 7

-(x - 5) = 2 -x + 5 = 2 x = 3

Verfifer:

| x - 5 | = 2| 7 - 5 | = 2 | 2 | = 2 2 = 2

| x - 5 | = 2| 3 - 5 | = 2 | -2 | = 2 2 = 2

Solving Absolute Value Equations Algebraically

a) | x + 1 | = x - 1 +(x + 1) = x - 1 x + 1 = x - 1 1 ≠ -1

-(x + 1) = x - 1 -x - 1 = x - 1 -2x = 0 x = 0

Pas de solution

b) | x - 2 | - | 2x + 11 | = 0

| x - 2 | = | 2x + 11 | (x - 2) = (2x + 11) x - 2 = 2x + 11 -x = 13 x = -13

-(x - 2) = (2x + 11) -x + 2 = 2x + 11 -3x = 9 x = -3

Algebriquement

a) | x + 1 | = x - 1

Graphiquement

b) | x - 2 | - | 2x + 11 | = 0

x = -13x = -3

Graphiquement

| x - 3| + | x - 8| = 17

(x - 3) + (x - 8) = 17 2x - 11 = 17 2x = 28 x = 14

- (x - 3) + -(x - 8) = 17 -x + 3 - x + 8 = 17 -2x + 11 = 17 -2x = 6 x = -3

-(x - 3) + (x - 8) = 17 -x + 3 + x - 8 = 17 -5 ≠ 17

(x - 3) + -(x - 8) = 17 x - 3 - x + 8 = 17 5 ≠ 17

Résoudre Algebriquement et graphiquement

| x - 3| + | x - 8| = 17

x = -3 ou x = 14

Devoir:

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