Răspuns :
[tex]x^2+y^2-4x-6y+13=0\\
(x^2-4x+4)+(y^2-6y+9)=0\\
(x-2)^2+(y-3)^2=0\\
Dar:(x-2)^2+(y-3)^2\geq 0,ceea\ ce\ implica\ faptul\ ca:\\
\{ x-2=0\Rightarrow \{x=2\\
\{y-3=0\Rightarrow \{y=3\\
(x-y)^{2016}+(y-x)^{2016}=(2-3)^{2016}+(3-2)^{2016}=\\
=(-1)^{2016}+1^{2016}=\bold{2}\\
[/tex]
..
x² + y² — 4x — 6y + 13 = 0
x² — 4x + 4 + y² — 6y + 9 = 0
( x — 2 )² + ( y — 3 )² = 0
x — 2 = 0
x = 2
y — 3 = 0
y = 3
( x — y )²⁰¹⁶ + ( y — x )²⁰¹⁶ → ( 2 — 3 )²⁰¹⁶ + ( 3 — 2 )²⁰¹⁶ = ( – 1 )²⁰¹⁶ + 1²⁰¹⁶ = 2
