Question

Tìm giá trị của biểu thức sau biết rằng a+b+c=0

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

Answer 1

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

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

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

\(=1+\dfrac{c}{a-b}.\dfrac{c\left(a-b\right)-\left(a+b\right)\left(a-b\right)}{ab}\)

\(=1+\dfrac{c}{a-b}.\dfrac{\left(c-a-b\right)\left(a-b\right)}{ab}\)

\(=1+\dfrac{c^2-\left(ab+ac\right)}{ab}=1+\dfrac{c^2-c\left(a+b\right)}{ab}\)

\(a+b+c=0\Leftrightarrow a+b=-c\)

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

Chứng minh tương tự và cộng theo vế,sử dụng khi \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\),suy ra đpcm