Question

[♫ Love Music ♫]

tìm x;y \(\in N\)

\(25-y^2=8\left|x-2009\right|\)

Answer 1

[Serena chuchoe]

Có: \(25-y^2\le25\)

\(\Rightarrow8\left|x-2009\right|\le25\)

\(\Rightarrow\left|x-2009\right|\le3\)

\(\Rightarrow\left[{}\begin{matrix}\left|x-2009\right|=3\\\left|x-2009\right|=2\\\left|x-2009\right|=1\\\left|x-2009\right|=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-2009=3\\x-2009=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2012\\x=2006\end{matrix}\right.\\\left[{}\begin{matrix}x-2009=2\\x-2009=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2011\\x=2007\end{matrix}\right.\\\left[{}\begin{matrix}x-2009=1\\x-2009=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2010\\x=2008\end{matrix}\right.\\x-2009=0\Rightarrow x=2009\end{matrix}\right.\)

=> Ta có các TH sau:

\(\left[{}\begin{matrix}25-y^2=8\cdot3=24\\25-y^2=8\cdot2=16\\25-y^2=8\cdot1=8\\25-y^2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y^2=1\\y^2=9\\y^2=17\\y^2=25\end{matrix}\right.\)

Vì y thuộc N nên: \(\left[{}\begin{matrix}y=1\\y=3\\y=\sqrt{17}\left(loai\right)\\y=5\end{matrix}\right.\)

=> các gt x;y thỏa mãn đề là:

\(\left[{}\begin{matrix}y=1\\y=3\\y=5\end{matrix}\right.\) lần lượt các gt x tương đương là\(\left[{}\begin{matrix}\left[{}\begin{matrix}x=2012\\x=2006\end{matrix}\right.\\\left[{}\begin{matrix}x=2011\\x=2007\end{matrix}\right.\\x=2009\end{matrix}\right.\)

Answer 2

[Linh_Windy]

Mk ko hiểu cách của bạn lắm :)

Theo đề bài: \(25-y^2=8\left|x-2009\right|\)

\(\left\{{}\begin{matrix}8\left|x-2009\right|\ge0\\8\left|x-2009\right|⋮8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}25-y^2\ge0\\25-y^2⋮8\end{matrix}\right.\)

Suy ra:\(y^2\le25\)

Vì \(y\in N\) nên \(y\) có thể là: \(\left\{0;1;2;3;4;5\right\}\)

Xét từng trường hợp ta có:

\(25-0^2=25⋮̸8\)

\(25-1^2=24⋮8\)

\(25-2^2=21⋮̸8\)

\(25-3^2=16⋮8\)

\(25-4^2=9⋮̸8\)

\(25-5^2=0⋮8\)

Vậy ta sẽ xét:

\(y=\left\{1;3;5\right\}\)

Xét lần lượt ta có:

\(\left[{}\begin{matrix}y=1\Rightarrow8\left|x-2009\right|=24\Rightarrow\left|x-2009\right|=3\Rightarrow\left[{}\begin{matrix}x=2012\\x=2006\end{matrix}\right.\\y=3\Rightarrow8\left|x-2009\right|=16\Rightarrow\left|x-2009\right|=2\Rightarrow\left[{}\begin{matrix}x=2011\\x=2007\end{matrix}\right.\\y=5\Rightarrow8\left|x-2009\right|=0\Rightarrow x=2009\end{matrix}\right.\)