đề bài thiếu nhé , a,b,c khác nhau nhé :)
có :\(a=b-c\)
vì a,b,c khác nhau
\(\Rightarrow b-c\ne0\)
có:
\(a+b+c=0\Leftrightarrow c=a-b.\)
\(a=b-c\)
\(b=c-a\)
thày vào M ta được
\(\left(\frac{c}{c}+\frac{a}{a}+\frac{b}{b}\right)\left(\frac{c}{c}+\frac{a}{a}+\frac{b}{b}\right)=9\)
\(M=\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)\)
Đặt \(\left(\frac{a-b}{c};\frac{b-c}{a};\frac{c-a}{b}\right)\rightarrow\left(x;y;z\right)\)
Khi đó \(M=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(=3+\frac{x+y}{z}+\frac{y+z}{x}+\frac{x+z}{y}\)
Ta có:\(\frac{x+y}{z}=\left(\frac{a-b}{c}+\frac{b-c}{a}\right)\cdot\frac{b}{c-a}\)
\(=\frac{a^2-ab+bc-c^2}{ca}\cdot\frac{b}{c-a}\)
\(=\frac{\left(a-c\right)\left(a+c\right)-b\left(a-c\right)}{ca}\cdot\frac{b}{c-a}\)
\(=\frac{\left(a-c\right)\left(a-b+c\right)}{ca}\cdot\frac{b}{c-a}\)
\(=\frac{-\left(a-b+c\right)\cdot b}{ca}=\frac{-\left(a+b+c-2b\right)\cdot b}{ca}=\frac{2b^3}{abc}\)
Tương tự ta có:
\(\frac{y+z}{x}=\frac{2c^3}{abc};\frac{x+z}{y}=\frac{2a^3}{abc}\)
Khi đó:\(M=\frac{2a^3+2b^3+2c^3}{abc}+3\)
Mặt khác:\(a+b+c=0\Rightarrow a+b=-c\Rightarrow a^3+3ab\left(a+b\right)+b^3=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó \(M=\frac{6abc}{abc}+3=9\)
