Question

Dacă numerele reale nenule x, y, z verifică relațiile $x+ \frac{1}{y}=1,$
$y+ \frac{1}{z}=2$ si $z+ \frac{1}{x}=5,$ atunci valoarea sumei x+y+z este egală cu:
A) 19/6; B) 23/6; C) 25/6; D) 29/6; E)6

Answer 1

x+1/y=1    x=1-1/y    
y+1/z=2    y=2-1/z =(2z-1) /z
x=1-1/[(2z-1) /z]=1-z/(2z-1) =(2z-1-z)/(2z-1)=(z-1)/(2z-1)
z+1/x=5
z+1/[(z-1)/(2z-1)]=5
z+(2z-1)/(z-1)=5
z²-z+2z-1=5z-5
z²-4z+4=0
(z-2)²=0
z-2=0
z=2
x=(z-1)/(2z-1)=(2-1)/(2*2-1)=1/3
y=(2z-1) /z=(2*2-1)/2=3/2

x+y+z=1/3+3/2+2=(2+9+12)/6=23/6
raspuns corect B)