[tex]\rm 4x^2+9y^2-4x+12y+5=0 \\ (4x^2-4x+1)+(9y^2+12y+4)=0 \\ (2x-1)^2+(3y+2)^2=0. \\ \\ (2x-1)^2 \geq 0 \\ (3y+2)^2 \geq 0 \\ \\ Deci ~(2x-1)^2+(3y+2)^2 \geq 0,~nsa~(2x-1)^2+(3y+2)^2=0. \\ \\ Egalitatea~are~loc~cand~(2x-1)^2~si~(3y+2)^2~au ~valoare~minima. \\ \\ Deci~(2x-1)^2=0 \Rightarrow 2x-1=0 \Rightarrow x= \frac{1}{2}. \\ (3y+2)^2=0 \Rightarrow 3y+2=0\Rightarrow y=-\frac{2}{3}. \\ \\ Solutie:~(x,y)=( \frac{1}{2};- \frac{2}{3}) [/tex]
