Question

Aflati nr reale x,y,z stiind ca x patrat+ypatrat+9zpatrat+2x-4y-12z+9=0

Answer 1

$\displaystyle x^2+y^2+9z^2+2x-4y-12z+9=0 \\ \\ x^2+2x+1+y^2-4y+4+9z^2-12z+4=0 \\ \\ (x+1)^2+(y-2)^2+(3z-2)^2=0 \\ \\ Patratele \ perfecte \ sunt \ mai \ mari \ sau \ egale \ cu \ 0 \ , \ iar \\ \\ suma \ patratelor \ noastre \ este \ 0\implies patratele \ sunt \ egale \ cu \ 0. \\ \\ (x+1)^2=0 \\ \\ x+1=0 \\ \\ x=-1 \\ \\ (y-2)^2=0 \\ \\ y-2=0 \\ \\ y=2 \\ \\ (3x-2)^2=0 \\ \\ 3z-2=0 \\ \\ z=\dfrac{2}{3}$