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Find the sum of x x x of all possible values of triples of positive real values (x,y,z) (x,y,z) (x,y,z) such that:
5(x+1x)=12(y+1y)=13(z+1z) 5( x+ \frac{1}{x}) = 12(y + \frac{1}{y}) = 13(z+ \frac{1}{z}) 5(x+x1)=12(y+y1)=13(z+z1)
xy+yz+zx=1xy+yz+zx = 1 xy+yz+zx=1
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