Có:\(a+b+c=0\Rightarrow c=-a-b\)
\(\Rightarrow b=-a-c\)
\(\Rightarrow a=-b-c\)
Cho \(B=\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\)
\(1\)\(\Rightarrow B.\dfrac{c}{a-b}=\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right).\dfrac{c}{a-b}\)
\(\Rightarrow B.\dfrac{c}{a-b}=1+\dfrac{c}{a-b}\left(\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\)
\(\Rightarrow B.\dfrac{c}{a-b}=1+\dfrac{c}{a-b}.\dfrac{b^2-bc+ac-a^2}{ab}\)
\(\Rightarrow B.\dfrac{c}{a-b}=1+\dfrac{c}{a-b}.\dfrac{\left(a-b\right)\left(c-a-b\right)}{ab}\)
\(\Rightarrow B.\dfrac{c}{a-b}=1+\dfrac{2c^2}{ab}=1+\dfrac{2c^3}{abc}\)
\(2\ B.\dfrac{a}{b-c}=\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right).\dfrac{a}{b-c}\)
\(\Rightarrow B\dfrac{a}{b-c}=1+\dfrac{a}{b-c}\left(\dfrac{a-b}{c}+\dfrac{c-a}{b}\right)\)
\(\Rightarrow B.\dfrac{a}{b-c}=1+\dfrac{a}{b-c}\left(\dfrac{ab-b^2+c^2-ac}{bc}\right)\)
\(\Rightarrow B.\dfrac{a}{b-c}=1+\left(\dfrac{\left(b-c\right)\left(b+c\right)-a\left(b+c\right)}{bc}\right)\dfrac{a}{b-c}\)
\(\Rightarrow B.\dfrac{a}{b-c}=1+\left(\dfrac{\left(b+c\right)\left(b-c-a\right)}{bc}\right).\dfrac{a}{b-c}\)
\(\Rightarrow B.\dfrac{a}{b-c}=1+\dfrac{2a^3}{abc}\)
\(3\ \)\(B.\dfrac{b}{c-a}=1+\dfrac{2b^3}{abc}\)
\(\Rightarrow A=\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\left(\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}\right)\)
\(\Rightarrow\left(B.\dfrac{c}{a-b}\right)+\left(B.\dfrac{a}{b-c}\right)+\left(B.\dfrac{b}{c-a}\right)\)
\(\Rightarrow1+\dfrac{2a^3}{abc}+1+\dfrac{2b^3}{abc}+1+\dfrac{2c^3}{abc}\)
\(\Rightarrow3+\dfrac{2\left(a^3+b^3+c^3\right)}{abc}\)
Áp dụng hằng đẳng thức mở rộng \(a^3+b^3+c^3=3abc\) khi \(a+b+c=0\)
\(\Rightarrow A=9\)
nhailaier you ngáo ak
Cho \(A=a+b+c=0\)
Tính \(A=\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\left(\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}\right)\) chắc chắn bằng 0 rồi :V
