Question

se dau nr rationale x, y, z astfel incat x + 2pe x = y + 3pe y = z + 4pe z = 144 calculati x, y, z

Answer 1

$Fie \ orice \ x, \ y, \ z\in Q \\ \frac{x+2}{x} = \frac{x+3}{y} = \frac{x+4}{z} =144 \\ x+2=144x \\ 2=144x-x \\ 2=143x \\ \boxed{x= \frac{2}{143} \approx\bold{0,014}} \\ \frac{0,014+3}{y} =144 \\ 144y=3,014 \\ \boxed{y= \frac{3,014}{144} \approx \bold{0,021}} \\ \frac{0,014+4}{z} =144 \\ 144z=4,014 \\ \boxed{z= \frac{4,014}{144}\approx \bold{0,028}}$