Question

Rút gọn biểu thức:

\(\dfrac{a}{\left(a-b\right)\left(a-c\right)}+\dfrac{b}{\left(b-c\right)\left(b-a\right)}+\dfrac{c}{\left(c-a\right)\left(c-b\right)}\) trong đó các số a,b,c là các số đôi một khác nhau

Answer 1

\(\dfrac{a}{\left(a-b\right)\left(a-c\right)}+\dfrac{b}{\left(b-c\right)\left(b-a\right)}+\dfrac{c}{\left(c-a\right)\left(c-b\right)}=\dfrac{a}{\left(a-b\right)\left(a-c\right)}-\dfrac{b}{\left(b-c\right)\left(a-b\right)}+\dfrac{c}{\left(a-c\right)\left(b-c\right)}=\dfrac{a\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\dfrac{b\left(a-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\dfrac{c\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\dfrac{ab-ac}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\dfrac{ab-bc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\dfrac{ac-bc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\dfrac{ab-ac-ab+bc+ac-bc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\dfrac{0}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)