Question

[Lê Hoàng Phúc]

cho a+b+c =0. Tính: \([\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}][\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)

Answer 1

[zZz Cool Kid_new zZz]

Đặt \(\left(\frac{a-b}{c};\frac{b-c}{a};\frac{c-a}{b}\right)\rightarrow\left(x;y;z\right)\)

Khi đó:

\(S=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\)

Ta có:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-cb+ac-a^2}{ab}\cdot\frac{c}{a-b}\)

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

\(=\frac{2c^2}{ab}=\frac{2c^3}{abc}\)

Một cách tương tự khi đó:\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}=\frac{2\left(a^3+b^3+c^3\right)}{abc}=\frac{2\cdot3abc}{abc}=6\)

Khi đó:\(S=3+6=9\) Bạn để ý rằng \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\)

Answer 2

[Lê Hoàng Phúc]

sao \(\frac{c\left(b+a-c\right)}{ab}\) lại bằng \(\frac{2c^2}{ab}\)