[tex]\rm 4 x^{2} +9 y^{2} -4x+12y+5=0 \\ \\ (4x^2-4x+1)+(9y^2+12y+4)=0 \\ \\ \underbrace{(2x-1)^2}+\underbrace{(3y+2)^2}=0 \\ ~~~\geq 0~~~~~~~~~~ \geq 0 \Longrightarrow (2x-1)^2+(3y+2)^2 \geq 0,~ \\ \\ dar~(2x-1)^2+(3y+2)^2=0 \Rightarrow (2x-1)^2=0 ~si~(3y+2)^2=0. \\ \\ (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} . \\ \\ \underline{Solutie}: ~ (a,b)= (\frac{1}{2} ;- \frac{3}{2} ).[/tex]
