Question

Tìm x; y, biết: \(x^2-y^2+2x-4y-10=0\) với x, y nguyên dương.

Answer 1

\(x^2-y^2+2x-4y-10=0\)

\(\Leftrightarrow x^2+2x+1-y^2-4y-4-7=0\)

\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=16\\\left(y+2\right)^2=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+1=4\\x+1=-4\left(l\right)\end{matrix}\right.\\\left[{}\begin{matrix}y+2=3\\y+2=-3\left(l\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

Vậy x = 3; y = 1.