Question

\(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)

Answer 1

Sửa đề: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)

Ta có: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)

\(=\dfrac{c-a}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\dfrac{a-b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\dfrac{b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\dfrac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)