Mathématiques

Question

Bonjour svp vous pouvez m’aider pour développer et réduire ces calculs mais pas factoriser ?
B=3x-(3-3y)+3(z+1)

C=(x+y-1)+(x+y+1)-(x+y-z)

D=2(x-y)+3(y-z)+4(z-x)+3x

E=x(1-y)+y(1-z)+z(1-x)+xy+yz+xz

F=x-[(1-y)-(z+1)]

2 Réponse

  • bonjour,


    B=3x-(3-3y)+3(z+1)
    = 3x-3+3y+3z+3
    = 3x+3y+3z

    C=(x+y-1)+(x+y+1)-(x+y-z)
    = x+y-1+x+y+1 -x-y+z
    = 2x-x+2y-y+z
    = x+y+z

    D=2(x-y)+3(y-z)+4(z-x)+3x
    = 2x-2y+3y-3z+4z-4x+3x
    = x+y+z

    E=x(1-y)+y(1-z)+z(1-x)+xy+yz+xz
    = x-xy+y-yz+z-xz+xy+yz+xz
    = x+y+z

    F=x-[(1-y)-(z+1)]
    = x-[1-y-z-1]
    = x-1+y+z+1
    = x+y+z

  • B=3x-(3-3y)+3(z+1)= 3x-3+3y+3z+3= 3x+3y+3z
    C=(x+y-1)+(x+y+1)-(x+y-z)= x+y-1+x+y+1 -x-y+z= 2x-x+2y-y+z= x+y+z
    D=2(x-y)+3(y-z)+4(z-x)+3x= 2x-2y+3y-3z+4z-4x+3x= x+y+z
    E=x(1-y)+y(1-z)+z(1-x)+xy+yz+xz = x-xy+y-yz+z-xz+xy+yz+xz= x+y+z
    F=x-[(1-y)-(z+1)]= x-[1-y-z-1] = x-1+y+z+1= x+y+z

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