\(\dfrac{a^2}{a^2-b^2-c^2}=\dfrac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}=\dfrac{a^2}{\left(a-b\right)\left(-c\right)-c^2}=\dfrac{a^2}{c\left(b-a-c\right)}=\dfrac{a^2}{2bc}\\ \Leftrightarrow M=\sum\dfrac{a^2}{a^2-b^2-c^2}=\sum\dfrac{a^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}\\ \Leftrightarrow M=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{2abc}=0\)

\(a+b=-c\Leftrightarrow\left(a+b\right)^3=-c^3\)

\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=3abc\)

\(A=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)

làm thế nào để tang điểm hỏi đáp

Có a + b + c = 0

=> a + b = - c

=> (a + b)² = c²

=> a² + b² + 2ab = c²

=> a² + b² - c² = - 2ab

Tương tự, b² + c² - a² = - 2bc và c² + a² - b² = - 2ca

Do đó \(A=\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ca}=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}=-\frac{3}{2}\)

a+b+c=0=>a+b=-c=>a²+b²+2ab=c²=>a²+b²-c²=-2ab

Tương tự b²+c²-a²=-2bc,c²+a²-b²=-2ac

=>\(A=\frac{-ab}{2ab}+\frac{-bc}{2bc}+\frac{-ca}{2ca}=\frac{-3}{2}\)

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