Răspuns:
(x,y,z)=(1,2,6),(1,3,5),(2,3,4)
Explicație pas cu pas:
x <y <z si x,(yz)+ y,(zx)+z,(xy) = x+ y + z + 1 |-x-y-z, ⇒ 0,(yz)+0,(zx)+0,(xy)=1, ⇒
[tex]\dfrac{yz}{99} +\dfrac{zx}{99}+\dfrac{xy}{99}=1,~\dfrac{yz+zx+xy}{99}=1~deci~ yz+zx+xy=99.\\[/tex]
10·y+z+10·z+x+10·x+y=99, ⇒11x+11y+11z=99, ⇒11·(x+y+z)=99, ⇒x+y+z=99:11, ⇒x+y+z=9. Deoarece x<y<z, rezulta
(x,y,z)=(1,2,6),(1,3,5),(2,3,4).
