a,b,c phân biệt \(\Rightarrow a\ne b\ne c\)
\(a^2\left(b+c\right)=b^2\left(c+a\right)=2012\)
\(\Rightarrow a^2b-ab^2+a^2c-b^2c=0\)
\(\Rightarrow ab\left(a-b\right)+c\left(a-b\right)\left(a+b\right)=0\)
\(\Rightarrow\left(a-b\right)\left(ab+bc+ca\right)=0\)
\(\Rightarrow ab+bc+ca=0\) vì \(a\ne b\)
\(a^2\left(b+c\right)=b^2\left(c+a\right)\)
\(\Rightarrow\dfrac{a^2}{a+c}=\dfrac{b^2}{b+c}=\dfrac{a^2-b^2}{a-b}=a+b\)
\(\Rightarrow a^2=\left(a+b\right)\left(a+c\right)\)
\(\Rightarrow2012=a^2\left(b+c\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(a+b\right)\left(ab+bc+ca+c^2\right)=c^2\left(a+b\right)\)
Vậy....................
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