A) 0,x(y)=0,y(5)=0,(5)
[tex] \frac{xy-x}{90}= \frac{y5-y}{90}= \frac{5}{9} [/tex]
[tex] \frac{10x+y-x}{90}= \frac{10y+5-y}{90}= \frac{50}{90} [/tex]
[tex] \frac{9x+y}{90}= \frac{9y+5}{90}= \frac{50}{90} [/tex]
Atunci 9x+y=50 si 9y+5=50
9y+5=50=> 9y=45 => y=5
9x+y=50 => 9x+5=50 => 9x=45 => x=5
B)
0,xy(z)+0,yz(x)+0,xz(y)=0,(5)
[tex] \frac{xyz-xy}{900} + \frac{yzx-yz}{900} + \frac{xzy-xz}{900} = \frac{5}{9} [/tex]
[tex] \frac{100x+10y+z-10x-y}{900} + \frac{100y+10z+x-10y-z}{900} + \frac{100x+10z+y-10x-z}{900} = \frac{500}{900} [/tex]
[tex] \frac{90x+9y+z}{900} + \frac{90y+9z+x}{900} + \frac{90x+9z+y}{900} = \frac{500}{900} [/tex]
Rezulta ca
90x+9y+z+90y+9z+x+90x+9z+y=500 =>
(90+1+90)*x+(9+90+1)*y+(1+9+9)*z=500
181x+100y+19z=500, unde x,y,z cifre
Daca x=0, 100y+19z=500 19z tb sa fie divizibil cu 100 deci z=0, y=5
Daca x=1, 100y+19z=319 => y=3, z=1
Daca x=2, 100y+19z=138 => y=1, z=2
Daca x=3, 100y+19z=-43, imposibil
Deci variantele sunt
x=0, y=5, z=0
x=1, y=3, z=1
x=2, y=1, z=2
