Question

giúp mình với 

Answer 1

Ta có: \(25-y^2\le25\)

\(\Rightarrow8\left(x-2018\right)^2\le25\)

\(\Rightarrow\left(x-2018\right)^2\le\dfrac{25}{8}\)

Mà \(\left(x-2018\right)^2\) là SCP

\(\Rightarrow\left(x-2018\right)^2=\left\{0;1\right\}\)

TH1: \(\left(x-2018\right)^2=0\)

\(\Rightarrow25-y^2=8.0=0\)

\(\Rightarrow y^2=25\Rightarrow\left[{}\begin{matrix}y=5\\y=-5\end{matrix}\right.\)

TH2: \(\left(x-2018\right)^2=1\)

\(\Rightarrow25-y^2=8.1=8\)

\(\Rightarrow y^2=17\Rightarrow\left[{}\begin{matrix}y=\sqrt{17}\\y=-\sqrt{17}\end{matrix}\right.\)

mà \(x,y\in N\)

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