Ta có :
\(25-y^2=8\left(x-2009\right)^2\)
\(\Rightarrow8\left(x-2009\right)^2\le25\)
\(\Leftrightarrow\left(x-2009\right)^2\le\frac{25}{8}\)
\(\Rightarrow0\le\left(x-2009\right)^2\le3\)
\(\Rightarrow\left(x-2009\right)^2\in\left\{0;1\right\}\)
+) Trường hợp 1 :
\(\Rightarrow\left(x-2009\right)^2=0\)
\(\Rightarrow x=2009\)
\(\Rightarrow y=5\)
\(\Leftrightarrow\hept{\begin{cases}x=2009\\y=5\end{cases}}\)
+) Trường hợp 2 :
\(\left(x-2009\right)^2=1\)
\(\Rightarrow x-2009=1\)
\(\Rightarrow x=2010\)
\(\Rightarrow25-y^2=8\)
\(\Rightarrow y^2=17\) (loại)
+) Trường hợp 3 :
\(\left(x-2009\right)^2=1\)
\(\Rightarrow x=2008\)
\(\Rightarrow25-y^2=8\)(loại)
Vậy ......
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