2)Ta có: \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Rightarrow2\left(ab+bc+ca\right)=-1\Rightarrow ab+bc+ca=-\dfrac{1}{2}\)
\(\Rightarrow\left(ab+bc+ca\right)^2=\dfrac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2\left(ab^2c+a^2bc+abc^2\right)=\dfrac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a=\dfrac{1}{4}\)
Mặt khác: \(a^2+b^2+c^2=1\Rightarrow\left(a^2+b^2+c^2\right)^2=1\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
\(\Rightarrow a^4+b^4+c^4=1-\dfrac{2.1}{4}=\dfrac{1}{2}\)
