\(\left(a+b+c\right)^2=0\) hay \(a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Rightarrow\left(2ab+2bc+2ac\right)=-1\)
\(\Rightarrow\left(2ab+2bc+2ac\right)^2=\left(-1\right)^2\)
\(4a^2b^2+4b^2c^2+4a^2c^2+8ab^2c+8abc^2+8a^2bc=1\)
\(2\left(2a^2b^2+2b^2c^2+2a^2c^2\right)+8abc\left(a+b+c\right)=1\)
\(\Rightarrow2\left(2a^2b^2+2b^2c^2+2a^2c^2\right)=1\)(vì a+b+c=0)
\(\Rightarrow2a^2b^2+2b^2c^2+2a^2b^2=1\)(1)
\(\left(a^2+b^2+c^2\right)^2=1\) hay \(a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=1\)
\(\Rightarrow a^4+b^4+c^4=1-\left(2a^2b^2+2b^2c^2+2a^2c^2\right)\)(2)
Thay (1) vào (2) Ta có:
\(a^4+b^4+c^4=\)\(1-\frac{1}{2}=\frac{1}{2}\) .
giúp mk vs ạ ai nhanh mk tick nha 
