Question

Chứng minh
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}\) + \(\frac{c-a}{\left(b-c\right)\left(b-a\right)}\) + \(\frac{a-b}{\left(c-a\right)\left(c-b\right)}\) = \(\frac{2}{a-b}\) + \(\frac{2}{b-c}\) + \(\frac{2}{c-a}\)

Answer 1

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

\(=\frac{-\left(b-c\right)^2-\left(c-a\right)^2-\left(a-b\right)^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{-2a^2-2b^2-2c^2+2ab+2ac+2bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\frac{2ab-2ac+2bc-2b^2+2ab+2ac-2bc-2a^2-2ab+2ac+2bc-2c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\frac{2\left(a-b\right)\left(b-c\right)+2\left(a-b\right)\left(c-a\right)+2\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\frac{2}{c-a}+\frac{2}{b-c}+\frac{2}{a-b}\)