Mathématiques

Question

4(1sur2y-1)(1sur2y+1) merci de m aider urgent

2 Réponse

  • 4(1/2y-1)(1/2y/1)=4(1/4y²+2y-2y-1)=4(1/4y²-1)=4/4y²-1

  •   4(1/2 y-1)(1/2 y+1 )

    =(4^1/2 y ) -(4^1) (1/2 y+1)

    =(2y -4 ) (1/2 y+1)

    =2y^1/2 y +2y ^1 -4^1/2 y -4^1

    =1y² +2y -2y -4

    =1y²-4

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