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 :)
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.
