Question

[vũ thị ánh dương]

Cho ,a,b,c khác 0 thỏa: \(a+b+c=0\) Đặt:

\(P=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b},Q=\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\)

Tính \(P.Q\)

giúp mk vs 

Answer 1

[Thanh Tùng DZ]

Đặt \(\frac{a-b}{c}=x;\frac{b-c}{a}=y;\frac{c-a}{b}=z\)\(\Rightarrow\frac{c}{a-b}=\frac{1}{x};\frac{a}{b-c}=\frac{1}{y};\frac{b}{c-a}=\frac{1}{z}\)

\(\Rightarrow P.Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\)

\(=3+\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}\)

Ta có : \(\frac{y+z}{x}=\left(y+z\right)\frac{1}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\frac{c}{a-b}=\left(\frac{b^2-bc+ac-a^2}{ab}\right)\frac{c}{a-b}\)

\(=\frac{\left(b-a\right)\left(a+b-c\right)}{ab}\frac{c}{a-b}=\frac{\left(c-a-b\right)c}{ab}=\frac{2c^2}{ab}\)( a + b + c = 0 suy ra c = -a-b )

Tương tự : \(\frac{x+z}{y}=\frac{2a^2}{bc};\frac{x+y}{z}=\frac{2b^2}{ac}\)

\(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}=\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=\frac{2\left(a^3+b^3+c^3\right)}{abc}=\frac{2.3abc}{abc}=6\)

( vì a + b + c = 0 . CM được a³ + b³ + c³ = 3abc )

\(\Rightarrow P.Q=3+6=9\)