Bonjour tout le monde :)

J'ai juste besoin qu'on m'aide a résoudre ces exercices en les détaillant svp ( je suis perdu ... ) :

Exercice 1 :

1) x²+2x+1=4x²-12+9

2) x²+25 = 10x

3) 5x-3/x-2 = 0

4) 1/x+1 = 3/x+3

Exercice 2 :

1) (x+2)(X-5) <ou= 0

2) 7(-x-1/2)(2/3x-4) <ou= 0

3) x(-5x4) > 0

4) (x+3)² >ou= 16

5) -3x+6/x-1 >ou= 0

6) 2x-5 > 4/2x-5

1

Réponses

2013-04-12T19:15:53+02:00

1) 3x² - 14x + 8 = 0 =>  3(x² - 14/3x + 8/3) = 0 => x² - 14/3x + 49/9 -25/9 = 0

 

x² - 14/3x + 49/9 -25/9 = 0 =>( x² - 14/3x + 49/9) -25/9 = 0 => (x-7/3)² = 25/9

                                                                                                 x - 7/3 = 5/3 => x = 12/3 = 4

                                                                                                 x - 7/3 =- 5/3 => x = 2/3

2)x²+25 = 10x => x² - 10 + 25 = 0 => (x-5)² = 0 => x = 5

3) 5x-3/x-2 = 0 => il faut x différent de 2

    (5x(x-2) - 3) = 0 => 5x² - 10x - 3 = 0 => 5(x² - 2x -3/5) = 0

=> x² - 2x + 1 -8/5 = 0 => (x-1)² = 8/5 et tu continues comme au 1)

4) attention il aut x dif de -1 et -3

produit en croix

x+3 = 3x + 3 => 2x = 0 => x = 0

Exercice 2

1) (x+2)(X-5) <ou= 0 tableau de signes

x           -2               5

       +      0       -      0        +      solution [-2;5]

2) 7(-x-1/2)(2/3x-4) <ou= 0

x        -1/2                 6

      -      0          +      0      -        solution    <--- ; -1/2] U [6;--->

3) x(-5x+4) > 0

 

x         0                    4/5

     -     0          +          0        -    solution ]0;4/5[

4) (x+3)² >ou= 16 => (x+3)²- 16 >=0 => (x+3+4)(x+3-4) >= 0 (x+7)(x-1) >=0

x          -7                   1

      +     0          -         0     +          solution <---;-7] U [1; --->

5) -3x+6/x-1 >ou= 0   x          1                    2

                                         -     |           +       0       -     solution ]1;2]

 

6) 2x-5 > 4/2x-5 => 2x-5 - 4/2x-5 > 0 => ((2x-5)² - 4)/(2x-5) > 0  => (2x-5-2)(2x-5+2)/(2x-5) > 0

(2x-7)(2x-3)/(2x-5)

x                         3/2                 5/2                7/2

(2x-7)       -                       -                     -        0        +

(2x-3)       -           0          +                     +                 +

(2x-5)       -                       -         0           +                   +

 

                -            0         +          |           -         0          +   solution ]3/2; 5/2[ U ]7/2; --->