Ta có :
\(P+1=x+y+xy+1=\left(x+1\right)\left(y+1\right)=\left(\frac{b^2+c^2-a^2}{2bc}+1\right)\left[\frac{a^2-\left(b-c\right)^2}{\left(b+c\right)^2-a^2}+1\right]\)
\(=\frac{b^2+2ab+c^2-a^2}{2bc}.\frac{a^2-\left(b-c\right)^2+\left(b+c\right)^2-a^2}{\left(b+c\right)^2-a^2}\)
\(=\frac{\left(b+c\right)^2-\left(b-c\right)^2}{2bc}=\frac{b^2+2bc+c^2-b^2+2bc-c^2}{2bc}=\frac{4bc}{2bc}=2\)
\(\Rightarrow P=2-1=1\)
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