\(\hept{\begin{cases}xy+x+y=3< =>xy+x+y+1=4< =>\left(x+1\right)\left(y+1\right)=4\left(1\right)\\yz+y+z=8< =>yz+y+z+1=9< =>\left(y+1\right)\left(z+1\right)=9\left(2\right)\\xz+x+z=15< =>xz+x+z+1=16< =>\left(x+1\right)\left(z+1\right)=16\left(3\right)\end{cases}}\)
Từ (1) , (2) và (3):
\(=>\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2=4.9.16=576=24^2\)
Do x,y,z dương =>(x+1)(y+1)(z+1)=24
từ (1)=>z+1=24:4=6=>z=5
từ (2)=>x+1=\(\frac{8}{3}\)=>x=\(\frac{5}{3}\)
từ (3)=>y+1=\(\frac{3}{2}\)=>y=\(\frac{1}{2}\)
\(=>P=x+y+z=5+\frac{5}{3}+\frac{1}{2}=\frac{43}{6}\)