Gọi 3 số đó là a,b,c (a,b,c là số tự nhiên) \(\left(a\ge b\ge c\right)\). Ta có:
\(a+b+c=abc\)
\(\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)
\(\Rightarrow1\le\dfrac{1}{c^2}+\dfrac{1}{c^2}+\dfrac{1}{c^2}=\dfrac{3}{c^2}\Leftrightarrow c^2\le3\Leftrightarrow c\in\left\{0;1\right\}\)
TH1 : \(c=0\Rightarrow a+b=0\Rightarrow a=b=0\)
TH2:
\(c=1\Rightarrow a+b+1=ab\\ \Leftrightarrow\left(a-1\right)\left(b-1\right)=2\)
Mà \(a-1\ge b-1\ge c-1=0\)
\(\Rightarrow a-1=2;b-1=1\Leftrightarrow a=3;b=2\)
Vậy \(\left(a,b,c\right)\in\left\{\left(0,0,0\right);\left(1,2,3\right);\left(1,3,2\right);\left(2,3,1\right);\left(2,1,3\right);\left(3,1,2\right);\left(3,2,1\right)\right\}\)
