Đáp án:
(x,y,z)=(4,3,1)
Lời giải:
$2x^2 + 2y^2 + z^2 + 25 – 6y- 2xy – 8x + 2z (y – x) = 0 $
$\Rightarrow x^2+y^2-2xy+2z(y-x)+z^2+x^2+y^2+25-6y-8x=0$
$\Rightarrow (y-x)^2+2z(y-x)+z^2+x^2-8x+16+y^2-6x+9=0$
$\Rightarrow (y-x+z)^2+(x-4)+(y-3)^2=0$
Do
$\left\{ \begin{array}{l} (y-x+z)^2\ge0 \\ (x-4)^2\ge0\\(y-3)^2\ge0 \end{array} \right .(\forall x)$
$\Rightarrow $ phương trình bằng 0 khi:
$\left\{ \begin{array}{l} (y-x+z)^2=0 \\ (x-4)^2=0\\(y-3)^2=0 \end{array} \right .$
$\Rightarrow \left\{ \begin{array}{l} y-x+z=0 \\ x-4=0\\y-3=0 \end{array} \right .$
$\Rightarrow \left\{ \begin{array}{l} z=x-y=4-3=1 \\ x=4\\y=3 \end{array} \right .$
