Question

Giải phương trình nghiệm nguyên

a) x²-4xy+5y²=169

b) 16x-25y=1

c)x².(x+2y)-y².(y+2x)=1991

Answer 1

\(a,x^2-4xy+5y^2=169\\ \Leftrightarrow\left(x-2y\right)^2+y^2=169\\ Vìx,y\in Znên:\\ \left[{}\begin{matrix}\left\{{}\begin{matrix}\left(x-2y\right)^2=0\\y^2=169\end{matrix}\right.\\\left\{{}\begin{matrix}\left(x-2y\right)^2=169\\y^2=0\end{matrix}\right.\\\left\{{}\begin{matrix}\left(x-2y\right)^2=25\\y^2=144\end{matrix}\right.\\\left\{{}\begin{matrix}\left(x-2y\right)^2=144\\y^2=25\end{matrix}\right.\end{matrix}\right.\\ Giảira\)