Question

Rút Gọn

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

biết a + b + c = 0

a khác b khác c khác 0

Answer 1

Vì  \(a+b+c=0\)  \(\Rightarrow\)  \(c=-a-b\)

Gọi  \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\) , ta có:

\(M.\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)=1+\frac{c}{a-b}.\frac{\left(b^2-bc+ac-a^2\right)}{ab}=1+\frac{c}{a-b}.\frac{\left(a-b\right)\left(c-a-b\right)}{ab}=1+\frac{2c^2}{ab}=1+\frac{2c^3}{abc}\)

Tương tự,  \(M.\frac{a}{b-c}=1+\frac{2a^3}{abc};\)  \(M.\frac{b}{c-a}=1+\frac{2b^3}{abc}\)

Mặt khác, ta cũng có: từ \(a+b+c=0\), suy ra \(a^3+b^3+c^3=3abc\)

Vậy,  \(B=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=3+6=9\)  (vì  \(a,b,c\ne0\)  nên \(abc\ne0\) )

Answer 2

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