Question

cho a+b+c=0 ,rút gọn :

B=\(\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{a^2-b^2-a^2}\)

Answer 1

Sửa đề:

\(\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-b^2-a^2}\)

Answer 2

ta có : \(B=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-b^2-a^2}\)

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

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

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

\(=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}\) \(=\dfrac{a^3+b^3+c^3}{2abc}\)