\(a+b+c=0=>a+b=-c=>\left(a+b\right)^2=\left(-c\right)^2=>a^2+2ab+b^2=c^2\)
\(=>a^2+2ab+b^2-c^2=0=>a^2+b^2-c^2=-2ab\)\(=>\left(a^2+b^2-c^2\right)^2=\left(-2ab\right)^2\)
\(=>a^4+b^4+c^4+2a^2b^2-2b^2c^2-2a^2c^2=4a^2b^2\)
\(=>a^4+b^4+c^4=4a^2b^2-\left(2a^2b^2-2b^2c^2-2a^2c^2\right)\)\(=2a^2b^2+2b^2c^2+2a^2c^2\)
\(=>2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=\left(a^2+b^2+c^2\right)^2=1^2\)\(=1\)
\(=>M=a^4+b^4+c^4=\frac{1}{2}\)
Ta có: \(a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^2=\left(-c\right)^2\)
\(\Rightarrow a^2+2ab+b^2=c^2\)
\(\Rightarrow a^2+2ab+b^2-c^2=0\)
\(\Rightarrow a^2+b^2-c^2=-2ab\)
\(\Rightarrow\left(a^2+b^2-c^2\right)^2=\left(-2ab\right)^2\)
\(\Rightarrow a^4+b^4+c^4+2a^2b^2-2b^2c^2-2a^2c^2=4a^2b^2\)
\(\Rightarrow a^4+b^4+c^4=4a^2b^2-\left(2a^2b^2-2b^2c^2-2a^2c^2\right)=2a^2b^2+2b^2c^2+2a^2c^2\)\(\Rightarrow2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=\left(a^2+b^2+c^2\right)^2=1^2\)\(\Rightarrow2\left(a^4+b^4+c^4\right)=1\)
\(\Rightarrow a^4+b^4+c^4=\dfrac{1}{2}\)
Vậy \(a^4+b^4+c^4=\dfrac{1}{2}\)
