Question

rezolvati sistemele
x(x+y+z)=6
y(x+y+z)=12
z(x+y+z)=18
rezolvarea completa daca se poate

Answer 1

Uite rezolvarea......

Answer 2

$x(x+y+z)=6\\ y(x+y+z)=12\\ z(x+y+z)=18\\ \\+ \\ (x+y+z)(x+y+z)=36 \\ (x+y+z)^{2} = 36\\(x+y+z)= \sqrt{36} \\ (x+y+z)=6 \\sau\\(x+y+z)=-6\\ 1. (x+y+z)=6\\ x(x+y+z)=6=\ \textgreater \ x=1\\ y(x+y+z)=12=\ \textgreater \ y=2 \\ z(x+y+z)=18=\ \textgreater \ z=3\\ \\ 2. (x+y+z)=-6\\ x(x+y+z)=6=\ \textgreater \ x=-1\\ y(x+y+z)=12=\ \textgreater \ y=-2\\ z(x+y+z)=18=\ \textgreater \ z=-3\\ S= {(1; 2; 3); (-1; -2; -3)}$