Question

Tìm cặp số nguyên ( x;y ) sao co ( x-5)^{2 +} ​​/2y + 4/ = 0

Answer 1

\(\left(x-5\right)^2+\left|2y+4\right|=0\)

Mà \(\left\{{}\begin{matrix}\left(x-5\right)^2\ge0\\\left|2y+4\right|\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left(x-5\right)^2+\left|2y+4\right|\ge0\)

Lại có : \(\left(x-5\right)^2+\left|2y+4\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-5\right)^2=0\\\left|2y+4\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\2y+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=-2\end{matrix}\right.\)

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