Question

[piojoi]

Tìm \(x,\) \(y\in Z\) biết: \(36-y^2=8(x-2010)^2\)

Answer 1

[Toru]

Ta có: \(y^2\ge0\forall y\in Z\)

\(\Rightarrow-y^2\le0\forall y\in Z\)

\(\Rightarrow36-y^2\le36\forall y\in Z\)

mà \(36-y^2=8\left(x-2010\right)^2\) (*)

nên \(8\left(x-2010\right)^2\le36\forall x\in Z\)

\(\Rightarrow\left(x-2010\right)^2\le\dfrac{36}{8}< 5\)

Mặt khác: \(\left(x-2010\right)^2\ge0\forall x\in Z\)

\(\Rightarrow\left(x-2010\right)^2\in\left\{0;1;2;3;4\right\}\)   (1)

Lại có: \(x\in Z\) nên \(x-2010\in Z\) (2)

Từ (1) và (2) \(\Rightarrow\left(x-2010\right)^2\in\left\{0;1;4\right\}\)

+, Với \(x-2010=0\Leftrightarrow x=2010\) , (*) trở thành:

\(36-y^2=0\)

\(\Rightarrow y^2=36\Rightarrow\left[{}\begin{matrix}y=6\\y=-6\end{matrix}\right.\left(tm\right)\)

+, Với \(\left(x-2010\right)^2=1\Leftrightarrow\left[{}\begin{matrix}x-2010=1\\x-2010=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2011\\x=2009\end{matrix}\right.\)

Khi đó: (*) ⇔ \(36-y^2=8\)

\(\Rightarrow y^2=28\Rightarrow y=\pm\sqrt{28}\left(ktm\right)\)

+, Với \(\left(x-2010\right)^2=4\Leftrightarrow\left[{}\begin{matrix}x-2010=2\\x-2010=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2010\\x=2008\end{matrix}\right.\)

Khi đó: (*) ⇔ \(36-y^2=8\cdot4\)

\(\Rightarrow y^2=4\Leftrightarrow\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\left(tm\right)\)

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