Question

tính tổng sau:

1/(b-c)(a^2+ac-b^2-bc)+1/(c-a)(b^2+ab-c^2-ac)+1/(a-b)(c^2+bc-a^2-ab)

Answer 1

\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)

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

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

\(=0\)