Question

Tính giá trị của biểu thức sau , biết rằng a+b+c=0 :

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

Answer 1

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

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

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

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

Vậy \(A=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=9\)