Question

Tìm x,y nguyên:

\(\left\{{}\begin{matrix}x+\dfrac{1}{\sqrt{y}}=2\\y+\dfrac{1}{\sqrt{x}}=2\end{matrix}\right.\)

Answer 1

ĐK: x,y >/ 0

hpt \(\Leftrightarrow\left\{{}\begin{matrix}x=2-\dfrac{1}{\sqrt{y}}\\y+\dfrac{1}{2-\dfrac{1}{\sqrt{y}}}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2-\dfrac{1}{\sqrt{y}}\\2y-y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\left(N\right)\\\left[{}\begin{matrix}y=1\left(N\right)\\y=-\dfrac{1}{2}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)

KL: x= y=1