Question

[Nguyễn Quang Tùng]

cho a+b+c=0 ( a khác 0 , b khác 0 , c khác 0)

tính giá trị biểu thức \(\frac{a^2}{a^2-b^2-c^2}\)+ \(\frac{b^2}{b^2-c^2-a^2}\)+\(\frac{c^2}{c^2-a^2-b^2}\)

Answer 1

[Ngô Thị Yến]

http://diendantoanhoc.net/topic/152549-t%C3%ADnh-fraca2a2-b2-c2-fracb2b2-c2-a2fracc2c2-b2-a2/

Answer 2

[alibaba nguyễn]

Ta có: \(a+b+c=0\)

\(\Rightarrow1\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a^2+b^2=-2ab+c^2\\b^2+c^2=-2bc+a^2\\c^2+a^2=-2ac+b^2\end{cases}}\)

\(\Rightarrow1A=\frac{a^2}{a^2+2bc-a^2}+\frac{b^2}{b^2+2ac-b^2}+\frac{c^2}{c^2+2ab-c^2}\)

\(=\frac{a^3+b^3+c^3}{2abc}=\frac{a^3+b^3+c^3-3abc+3abc}{2abc}\)

\(=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\)

\(=\frac{3}{2}\)