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1810moto
@1810moto
May 2019
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Il faut développer et factoriser mais je n'y arrive vrmt pas aidez moi svp c'est un dm
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nadiab
Bonjour,
A= 2x³+3x-2x²-3⇔ 2x³-2x²+3x-3
B= x²+2x+1+4x²-4x+1
B= réduis
C= 4x²+4x+1-(x²-4x+4)
C= 4x²+4x+1-x²+4x-4
C= réduis
Factoriser
A= (x+1)(4x-3+2x+1)
A= réduis
B= (2x+1)(2x+1+3x-1)
B= réduis
C= [(x-1)-(2x-3)](x-1+2x-3)
C= (x-1-2x+3)(3x-4)
C= réduis
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A= 2x³+3x-2x²-3⇔ 2x³-2x²+3x-3
B= x²+2x+1+4x²-4x+1
B= réduis
C= 4x²+4x+1-(x²-4x+4)
C= 4x²+4x+1-x²+4x-4
C= réduis
Factoriser
A= (x+1)(4x-3+2x+1)
A= réduis
B= (2x+1)(2x+1+3x-1)
B= réduis
C= [(x-1)-(2x-3)](x-1+2x-3)
C= (x-1-2x+3)(3x-4)
C= réduis