Question

4x^2+9y^2+12 (x-y)+13 <sau egal cu 0 . Să se afle x şi y

Answer 1

[tex]4x^2+9y^2+12(x-y)+13 \leq 0\\ 4x^2+9y^2+12x-12y+13\leq 0\\ (4x^2+12x+9)+(9y^2-12y+4)\leq 0\\ (2x+3)^2+(3y-2)^2\leq 0\\ \text{Deoarece }(2x+3)^2\geq 0\ \text{si}\ (3y-2)^2\geq 0~\text{inseamna ca:}\\ (2x+3)^2+(3y-2)^2\geq 0,\ \text{ceea ce implica faptul ca:} \\ \left \{ {{2x+3=0} \atop {3y-2=0}} \right.\Leftrightarrow x=-\dfrac{3}{2}; y=\dfrac{2}{3}[/tex]