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Salut je dois résoudre 5x+1-(x-4)+3=4(x+2)

 

1/2(2y-3)-11/3(y+3)=1/9(4y-2)

 

5(t+1)²=5t²-1

 

Merciiiii d'avance :)

Sagot :

5x+1-(x-4)+3=4(x+2)

5x+1-(x-4)+3-4(x+2) = 0

5x + 1 -x + 4 + 3 - 4x - 8 = 0

0 = 0

donc x a toutes les valeurs possibles comme solution.

 

1/2(2y-3)-11/3(y+3)=1/9(4y-2)

1/2(2y-3)-11/3(y+3)-1/9(4y-2) = 0

9/18(2y-3)-66/18(y+3)-2/18(4y-2) = 0

9(2y-3)-66(y+3)-2(4y-2) = 0

18y-27-66y-198-8y+4 = 0

(18-66-8)y = 27+198-4

 -56y = 221

y = - 221/56

 

5(t+1)²=5t²-1

5(t+1)²-5t²+1 = 0

5(t² - 2t +1)-5t²+1 = 0

5t² - 10t +5 -5t²+1 = 0

-10t = -6

t = 6/10

t = 3/5

 

en espérant t'avoir aidé.

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