2x2+y2+9=6x+2xy
=>2x2+y2+9-6x-2xy=0
=>(x2-2xy+y2)+(x2-6x+9)=0
=>(x-y)2+(x-3)2=0
do (x-y)2 ≥ 0 ∀ x,y
(x-3)2 ≥ 0 ∀x
=>(x-y)2+(x-3)2 =0 khi
=>\(\left[{}\begin{matrix}x-y=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=x=3\\x=3\end{matrix}\right.\)
thay x=3 và y=3
Q=32017.32018-32018. 32017+\(\dfrac{1}{9}.3.3\)
Q=1