Ta có :
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}\)
\(=\dfrac{1}{a\left(a-b\right)\left(a-c\right)}-\dfrac{1}{b\left(a-b\right)\left(b-c\right)}\)
\(=\dfrac{b\left(b-c\right)-a\left(a-c\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{b^2-bc-a^2+ac}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(b^2-a^2\right)-\left(bc-ac\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(b-a\right)\left(b+a\right)-c\left(b-a\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(b-a\right)\left(b+a-c\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(a-b\right)\left(-a-b+c\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{-a-b+c}{ab\left(a-c\right)\left(b-c\right)}\)
Như vậy :
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)
\(=\dfrac{-a-b+c}{ab\left(a-c\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)
\(=\dfrac{-a-b+c}{ab\left(a-c\right)\left(b-c\right)}+\dfrac{1}{c\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{c\left(-a-b+c\right)+ab}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{-ac-bc+c^2+ab}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(-ac+ab\right)-\left(bc-c^2\right)}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{a\left(b-c\right)-c\left(b-c\right)}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(a-c\right)\left(b-c\right)}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{1}{abc}\)