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bonjour j'ai besoin de votre aide merciii :
g(x)=(x−1) au carré −(x−1)
g(x)=(x−1) au carré −(x+1) au carré
g(x)=((x−1)+(x+1)) au carré
g(x)=x(x+1)−(x−1) au carré
merci d'avance et on ne connait pas le x

Sagot :

GGDU19

Bonjour,

g(x)=(x−1)²-(x−1)

     =(x-1)((x-1)-1)

g(x)=(x-1)(x-2)

g(x)=(x−1)²−(x+1)²

     =x²-2x+1-x²-2x-1

g(x)=-4x

g(x)=((x−1)+(x+1))²

     =(x-1+x+1)²

     =(2x)²

g(x)=4x²

g(x)=x(x+1)−(x−1)²

     =x²+x-x²+2x-1

f(x)=3x-1

Bonjour,

g(x)=(x−1) au carré −(x−1)

g (x) = (x - 1) [(x - 1) - 1]

g (x) = (x - 1) (x - 1 - 1)

g (x) = (x - 1) (x - 2)

g(x)=(x−1) au carré −(x+1) au carré

g (x) = x² - 2x + 1 - (x² + 2x + 1)

g (x) = x² - 2x + 1 - x² - 2x - 1

g (x) = x² - x² - 2x - 2x + 1 - 1

g (x) = - 4x

g(x)=((x−1)+(x+1)) au carré

g (x) = (x - 1 + x + 1)²

g (x) = (2x)²

g (x) = (4x)²

g(x)=x(x+1)−(x−1) au carré

g (x) = x² + x - (x² - 2x + 1)

g (x) = x² + x - x² + 2x - 1

g (x) = x² - x² + x + 2x - 1

g (x) = 3x - 1.

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