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\(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow14+2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=-7\)
Suy ra : \(\left(ab+bc+ac\right)^2=49\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)
\(a^2+b^2+c^2=14\Leftrightarrow\left(a^2+b^2+c^2\right)^2=196\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)
\(\Leftrightarrow a^4+b^4+c^4+2.49=256\) \(\Leftrightarrow a^4+b^4+c^4=98\)
Vậy ...
\(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc +2ca=0\)
\(\Leftrightarrow2ab+2bc+2ca=-14\)
\(\Leftrightarrow ab+bc+ca=-7\)
\(\Rightarrow\left(ab+bc+ca\right)^2=49\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc=49\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=49\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=49\).
\(a^2+b^2+c^2=14\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2=14^2=196\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=196\)
\(\Leftrightarrow a^4+b^4+c^4+2.49=196\)
\(\Leftrightarrow a^4+b^4+c^4=98\)
Ta có:
\(ab+bc+ca=\frac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}=\frac{0-2010}{2}=-1005\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=\left(ab+bc+ca\right)^2-2abc\left(a+b+c\right)\)
\(=\left(-1005\right)^2-2abc.0=1005^2\)
\(\Rightarrow A=a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(=2010^2-1005^2=2.1005^2=2020050\)
+) Ta có : \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow2\left(ab+bc+ca\right)=-2016\)
\(\Rightarrow\left(ab+bc+ca\right)^2=\left(-2013\right)^2\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=2013^2\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=2013^2\)( Do \(a+b+c=0\) )
+) Lại có : \(a^2+b^2+c^2=2016\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2=2016^2\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=2016^2\)
\(\Rightarrow a^4+b^4+c^4=2016^2-2.2013^2=-4040082\)
Hay : \(A=-4040082\)
Vậy \(A=-4040082\) với a,b,c thỏa mãn đề.