Question

Tìm nhiệm nguyên 

x²y+2xy+y=81=0

Answer 1

\(y\left(x^2+2x+1\right)=81\)

\(\Leftrightarrow y=\frac{81}{\left(x+1\right)^2}\Rightarrow\left(x+1\right)^2\inƯ\left(81\right)=\left\{1;9;81\right\}\)

(do \(\left(x+1\right)^2\ge0\),\(\left(x+1\right)^2\)là số chính phương vì x nguyên)

\(\left(x+1\right)^2\)	1	9	81
x	0;-2	2;-4	8;-10
y	81	9	1

Kết luận:........