1) (x+3)(x-5)+2(x+3)(x+6)=0

 

2) (x-5)(x+15)=(x-5)(x+7)

 

3) V5x + 13 = V3x + 6

 

4) V2x + V3 > 4x - 5V3

 

 

(V) =( racine carré )

 

merci d avance

Répondre :

(x+3) [(x-5)+2(x+6)]

(x+3)(x+5+2x+12)

(x+3)(3x+17)

1) (x+3)(x-5)+2(x+3)(x+6)=0 <=> 1) (x+3)(x-5+2x+12)=0 <=> (x+3)(3x+7)=0

soit x+3 = 0 <=> x1=-3

soit 3x+7 = 0 <=>  3x=-7 <=> x2 = -7/3

2) (x-5)(x+15)=(x-5)(x+7) <=> 2) (x-5)(x+15)-(x-5)(x+7)=0 

<=> 2) (x-5)(x+15-x-7)=0 <=> 8(x-5)=0 <=> x-5 = 0 <=> x = 5

3) V5x + 13 = V3x + 6 <=> (V5-V3)x = 6-13 = -7 <=> x= -7/(V5-V3)

4)  V2x + V3 > 4x - 5V3 <=>  (V2 - 4)x  > - 6V3 <=> x  > - (6V3)/(V2 - 4)

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