Đặt \(b^2+c^2-c^2=t;2bc=u\), ta có:
\(x=\frac{t}{u};y=\frac{a^2-b^2-c^2+2bc}{b^2+c^2-a^2+2bc}=\frac{2bc-\left(b^2+c^2-a^2\right)}{2bc+\left(b^2+c^2-a^2\right)}=\frac{u-t}{u+t}\)
nên \(P=x+y+xy=x+1+y\left(x+1\right)-1=\left(x+1\right)\left(y+1\right)-1\)
\(P=\left(\frac{t}{u}+1\right)\left(\frac{u-t}{u+t}+1\right)-1=\frac{t+u}{u}.\frac{2u}{u+1}-1=2-1=1\)