Factoriser:
9-12y+4y²

25y²+10y+1

4y²-49

7y²+5y

(3y-5)²-(3y-5)(12y+4)

(6y-5)(4y+7)+(6y-5)(2y-5)

(3y-2)(6y+7)-(3y-2)(-2y+4)

(6y-8)²-(2y-3)²

(5y-8)²+(5y-8)

(2y+5)²-81

2
je te donne les étapes ou seulement la réponse?

Réponses

2014-02-13T18:59:50+01:00
Bonsoir

9-12y+4y² = (3-2y)²

25y²+10y+1 = (5y +1)²

4y²-49 = (2y -7)(2y +7)

7y²+5y = (y) ( 7y +5)
(3y-5)²-(3y-5)(12y+4) = (3y -5)(3y-5) - (12y +4) = (3y-5)(3y-5 -12y-4) = (3y-5)(-9y -9)

(6y-5)(4y+7)+(6y-5)(2y-5) = (6y-5)(4y+7 + 2y-5) = (8y-5)(6y +2)

(3y-2)(6y+7)-(3y-2)(-2y+4) = (3y-2)(6y+7 +2y-4) = (3y-2)( 8y +3)

(6y-8)²-(2y-3)² = (6y-8 -2y +3)(6y-8 +2y -3) = (4y -5)(8y -11)

(5y-8)²+(5y-8) = (5y-8)(5y-8 +1) = (5y-8)(5y -7)

(2y+5)²-81
= (2y +5 +9)(2y +5 - 9) = (2y +14)(2y -4)
2014-02-13T19:05:34+01:00
9-12y+4y² = 3²-3*2*2y+(2y)² = (3-2y)²  identité remarquable 

25y²+10y+1= (5y+1)²  identité remarquable 

4y²-49 = (2y)²-7² = (2y-7)(2y+7)  identité remarquable 

7y²+5y = y(7y+5) 

(3y-5)²-(3y-5)(12y+4) = (3y-5)(3y-5-12y-4) = (3y-5)(-9y-9)  attention au changement de signes a cause du moins 

(6y-5)(4y+7)+(6y-5)(2y-5) = (6y-5)(4y+7+2y-5) = (6y-5)(6y+2)

(3y-2)(6y+7)-(3y-2)(-2y+4) = (3y-2)(6y+7+2y-4) = (3y-2)(8y+3)

(6y-8)²-(2y-3)² = (6y-8-2y+3)²= (4y-5)² la je ne suis pas sure 

(5y-8)²+(5y-8) = (5y-8)(5y-8+1) = (5y-8)(5y-7)
 
(2y+5)²-81 = (2y+5)²-9² = (2y+5-9)(2y+5+9) = (2y-4)(2y+14)

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