Ta có :
+) \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Rightarrow1+2\left(ab+bc+ac\right)=0\)
\(\Rightarrow2\left(ab+bc+ac\right)=-1\)
\(\Rightarrow ab+bc+ac=-\dfrac{1}{2}\)
+) \(ab+bc+ac=-\dfrac{1}{2}\)
\(\Rightarrow\left(ab+bc+ac\right)^2=\left(-\dfrac{1}{2}\right)^2\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+2abc.0=\dfrac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+0=\dfrac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=\dfrac{1}{4}\)
+) \(a^2+b^2+c^2=1\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2=1\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=1\)
\(\Rightarrow a^4+b^4+c^4+2.\dfrac{1}{4}=1\)
\(\Rightarrow a^4+b^4+c^4+\dfrac{1}{2}=1\)
\(\Rightarrow a^4+b^4+c^4=\dfrac{1}{2}\)
Vậy \(a^4+b^4+c^4=\dfrac{1}{2}\)
