x2 + 2y2 + 2xy - 4x + 6y + 29 = 0
<=> ( x2 + 2xy + y2 - 4x - 4y + 4 ) + ( y2 + 10y + 25 ) = 0
<=> [ ( x2 + 2xy + y2 ) - 2( x + y ).2 + 22 ] + ( y + 5 )2 = 0
<=> ( x + y - 2 )2 + ( y + 5 )2 = 0 (*)
<=> \(\hept{\begin{cases}\left(x+y-2\right)^2\ge0\forall x,y\\\left(y+5\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x+y-2\right)^2+\left(y+5\right)^2\ge0\forall x,y\)
Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}x+y-2=0\\y+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=7\\y=-5\end{cases}}\)
Vậy x = 7 ; y = -5
