$x+1=\frac{b^2+c^2-a^2}{2bc}+1=\frac{b^2+2bc+c^2-a^2}{2bc}=\frac{(b+c)^2-a^2}{2bc}$
Suy ra
$y(x+1)=\frac{a^2-(b-c)^2}{(b+c)^2-a^2}.\frac{(b+c)^2-a^2}{2bc}=\frac{a^2-(b-c)^2}{2bc}$
Do đó
$P=x+y+xy=x+y(x+1)=\frac{b^2+c^2-a^2}{2bc}+\frac{a^2-(b-c)^2}{2bc}=\frac{b^2+c^2-a^2+a^2-(b-c)^2}{2bc}=1$
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