\(25-y^2=8.\left(x-2009\right)^2\left(1\right)\)
\(\rightarrow25=8.\left(x-2009\right)^2+y^2\)
\(NX:y^2\ge0\forall y\in R\rightarrow8.\left(x-2009\right)^2\le25\)
\(\rightarrow\left(x-2009\right)^2\le3,125\)
\(\rightarrow\left[{}\begin{matrix}x-2009=0\\x-2009=1\end{matrix}\right.\rightarrow\left[{}\begin{matrix}x=2009\\x=2010\end{matrix}\right.\)
Thay x=2009 vào (1) ta dược:
\(25-y^2=8.\left(2009-2009\right)^2\)
\(25-y^2=8.0\)
\(25.y^2=0\)
\(y^2=25\)
\(\rightarrow\left[{}\begin{matrix}y=5\\y=-5\left(L\right)\end{matrix}\right.\)
Thay x=2010 vào (1) ta được:
\(25-y^2=8.\left(2010-2009\right)^2\)
\(25-y^2=8.1\)
\(25-y^2=8\)
\(y^2=17\left(L\right)\)
Vậy\(\left(x;y\right)=\left(2009;5\right)\)
