Ta có : \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=-\frac{1}{2}\)
\(\Leftrightarrow\left(ab+bc+ac\right)^2=\frac{1}{4}\Leftrightarrow\left(a^2b^2+b^2c^2+c^2a^2\right)+2abc\left(a+b+c\right)=\frac{1}{4}\Rightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)
Mặt khác : \(\left(a^2+b^2+c^2\right)^2=1\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
\(\Rightarrow a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+c^2a^2\right)\Rightarrow a^4+b^4+c^4=\frac{1}{2}\)
(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac=0 => 2ab+2bc+2ac= -1 =>ab+bc+ac=-1/2
=>(ab+bc+ac)^2=1/4=0.25 =>a^2b^2+b^2c^2+a^2c^2+2a^2bc+ab^2c+abc^2=0.25
=>a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c)=0.25
=>a^2b^2+b^2c^2+a^2c^2=0.25 =>2a^2b^2+2b^2c^2+2a^2c^2=0.5 (1)
Mà (a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=1 (2)
Thay (1) vào (2) =>a^4+b^4+c^4=1-0.5=0.5
Vậy M=0.5
a+b+c=0 nên(a+b+c)^2=0 suy ra a^2+b^2+c^2 +2(ab+bc+ca)=0
a^2+b^2+c^2=-[2(ab+bc+ca)]
(a^2+b^2+c^2)^2 = [2(ab+bc+ca)]^2( bình phương nên bỏ mất dấu âm)
a^4+b^4+c^4+2(a^2.b^2+b^2.c^2+c^2.a^2)=4(a^2.b^2+b^2.c^2+c^2.a^2) +8abc(a+b+c)
a^4+b^4+c^4=2(a^2.b^2+b^2.c^2+c^2.a^2)=1/2(vì 8abc(a+b+c)=0, bỏ 2(a^2.b^2+b^2.c^2+c^2.a^2) ở cả hai vế và a^2+b^2+c^2=1 nên (a^2+b^2+c^2)^2=1 do đó 4(a^2.b^2+b^2.c^2+c^2.a^2)=1)
