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6(x+1)/2 + 6(x+3)/3 = 6(9)
3(x+1) + 2(x+3) = 6(9)
3x+3+2x+6 = 54
5x=54-9
5x=45
x=45/5
x=9
Ahora sustituimos:
(x-1)^(x-8) = (9-1)^(9-8) = 8^1 = 8
2. Multiplicamos todo por 15 para eliminar denominadores:
15(x+2)/3 + 15(x-1)/5 = 15x
5(x+2) + 3(x-1) = 15x
5x+10+3x-3=15x
8x-15x=3-10
-7x=-7
x=-7/-7
x=1
Sustituimos ("^3√" significa "raíz cúbica"):
^3√(7x+1) = ^3√(7(1)+1) = ^3√(7+1) = ^3√8 = 2