Question

Tìm tất cả các số nguyên x\(\ge\)y\(\ge\)z\(\ge\)0 thỏa mãn

xyz + xy + yz + zx + x + y + z = 2011

Answer 1

Ta có:

\(xyz+xy+yz+zx+x+y+z=2011\)

\(\Leftrightarrow xy\left(z+1\right)+y\left(z+1\right)+x\left(z+1\right)+\) \(\left(z+1\right)=2012\)

\(\Leftrightarrow\left(z+1\right)\left(xy+y+x+1\right)=2012\)

\(\Leftrightarrow\left(z+1\right)\left[x\left(y+1\right)+\left(y+1\right)\right]=2012\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)=2012\)

Mà \(2012=1.2.2.503=503.4.1\)

\(\Leftrightarrow\left[{}\begin{matrix}x=502;y=1;z=1\\x=1005;y=1;z=0\\x=2011;y=0;z=0\end{matrix}\right.\)

Vậy...