Có: \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow2009+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow ab+bc+ac=-\frac{2009}{2}\)
\(\Leftrightarrow\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=\left(-\frac{2009}{2}\right)^2\)
\(\Leftrightarrow\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2=\left(-\frac{2009}{2}\right)^2\)
Mặt khác: \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(=a^4+b^4+c^4+2.\left(-\frac{2009}{2}\right)^2=2009^2\)
\(\Leftrightarrow a^4+b^4+c^4=2009^2-2.\left(-\frac{2009}{2}\right)^2=2009^2-2.\frac{2009^2}{2^2}=2009^2-\frac{2009^2}{2}\)
--Hà Phương--