\(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}-\frac{1}{p}\)
\(=\frac{2}{2p-2a}+\frac{2}{2p-2b}+\frac{2}{2p-2c}-\frac{2}{2p}\)
\(=\frac{2}{a+b+c-2a}+\frac{2}{a+b+c-2b}+\frac{2}{a+b+c-2c}-\frac{2}{a+b+c}\)
\(=\frac{2}{b+c-a}-\frac{2}{b+c+a}+\frac{2}{a-b+c}+\frac{2}{a+b-c}\)
\(=\frac{2\left(b+c+a\right)-2\left(b+c-a\right)}{\left(b+c-a\right)\cdot\left(b+c+a\right)}+\frac{2\left(a+b-c\right)+2\left(a-b+c\right)}{\left(a-b+c\right)\left(a+b-c\right)}\)
\(=\frac{4a}{\left(b+c-a\right)\left(b+c+a\right)}+\frac{4a}{\left(a-b+c\right)\left(a+b-c\right)}\)
\(=4a\left(\frac{1}{\left(b+c\right)^2-a^2}+\frac{1}{a^2-\left(b-c\right)^2}\right)=4a\cdot\frac{a^2-\left(b-c\right)^2+\left(b+c\right)^2-a^2}{\left\lbrack\left(b+c\right)^2-a^2\right\rbrack\left\lbrack a^2-\left(b-c\right)^2\right\rbrack}\)
\(=4a\cdot\frac{b^2+2bc+c^2-b^2+2bc-c^2}{\left(b+c-a\right)\left(b+c+a\right)\left(a-b+c\right)\left(a+b-c\right)}\)
\(=4a\cdot\frac{4bc}{\left(b+c-a\right)\left(b+c+a\right)\left(a-b+c\right)\left(a+b-c\right)}\)
\(=16\cdot\frac{abc}{\left(b+c-a\right)\left(b+c+a\right)\left(a-b+c\right)\left(a+b-c\right)}\)
\(\frac{abc}{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)
\(=\frac{16abc}{2p\left(2p-2a\right)\left(2p-2b\right)\left(2p-2c\right)}\)
\(=\frac{16abc}{\left(a+b+c\right)\left(b+c-a\right)\left(a-b+c\right)\left(a+b-c\right)}\)
Do đó: \(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}-\frac{1}{p}=\frac{abc}{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)
