\(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Leftrightarrow2\left(ab+bc+ac\right)=-6\)
\(\Leftrightarrow ab+bc+ac=-3\)
\(\Leftrightarrow\left(ab+bc+ac\right)^2=9\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2ab^2c+2abc^2+2a^2bc=9\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=9\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=9\)
Ta có: \(a^2+b^2+c^2=6\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=36\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=36\)
\(\Leftrightarrow a^4+b^4+c^4=18\)
Vậy...
Đúng 0
Bình luận (0)
