1/2 nhá

a+b+c = 0 <=> (a+b+c)^2 = 0

<=> 2(ab+bc+ca) = 0 - (a^2+b^2+c^2) = 0 - 1 = -1

<=> ab+bc+ca = -1/2

<=> (ab+bc+ca)^2 = 1/4

<=> a^2b^2+b^2c^2+c^2a^2 = 1/4 - 2abc.(a+b+c) = 1/4 - 0 = 1/4

Có : a^2+b^2+c^2 = 1

<=> (a^2+b^2+c^2) = 1

<=>  A = a^4+b^4+c^4 = 1 - 2.(a^2b^2+b^2c^2+c^2a^2) = 1 - 2.1/4 = 1/2

Vậy A = 1/2

k mk nha

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)=2+2\left(ab+bc+ac\right)\)

=> \(0=2+2\left(ab+bc+ac\right)\)=> \(ab+bc+ca=-1\)

=> \(\left(ab+bc+ac\right)^2=1\)

Mà \(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+2\left(ab^2c+a^2bc+abc^2\right)\)

                                             \(=a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+a^2c^2\)

=> \(a^2b^2+b^2c^2+c^2a^2=1\)

Mặt khác : \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

=> \(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

                                             \(=4-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

=> \(a^4+b^4+c^4=4-2=2\)

Ta có :

( a + b + c )² = a² + b² + c² + 2ab + 2 bc+ 2ac = 0

Mà a² + b ² + c² = 1 

=> 2ab + 2bc + 2ac = - 1

=> ab + bc + ac = \(\frac{-1}{2}\)

=> ( ab + bc + ac ) ² = a²b² + a²c² + b²c ² + 2ab²c + 2ac²b + 2a²bc = \(\left(\frac{-1}{2}\right)^2\)=\(\frac{1}{4}\)

=> a²b² + a²c² + b²c² + 2abc ( a + b +c ) = \(\frac{1}{4}\)

mà a + b + c =  0 => 2abc ( a +b +c ) = 0

=> a²b² + b²c² + c²a² = \(\frac{1}{4}\)

Ta có : ( a² + b² + c² )² = a⁴ + b⁴ + c⁴ + 2 ( a²b² + b²c² + c²a² ) = 1

=> a⁴ +b⁴ + c⁴  + 2. \(\frac{1}{4}\) = 1

=> a⁴ + b⁴ + c⁴ = 1 - \(\frac{1}{2}\)

=> a⁴ + b⁴ + c⁴ = \(\frac{1}{2}\)

Bài 1:

\(a^2+b^2+c^2=16\Rightarrow\left(a+b+c\right)^2-2ab-2bc-2ac=16\)\(\Leftrightarrow-2\left(ab+bc+ac\right)=16\Rightarrow ab+bc+ac=-8\)\(\Rightarrow\left(ab+bc+ac\right)^2=64\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=64\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=64\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=64\)

Ta có:

\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2a^2b^2-2b^2c^2-2a^2c^2\)\(=16^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=256-2.64=128\)

1. Ta có $a + b + c = 0$

\(\Rightarrow\) $( a + b + c)^2 = 0$

\(\Leftrightarrow\) $a^2+b^2 +c^2 +2ab+2bc+2ac = 0
\(\Leftrightarrow\) $a^2 + b^2 + c^2 = -2(ab+bc+ac)$

Thay $a^2 + b^2 + c^2 = 2$

\(\Rightarrow\)$2 = -2(ab+bc+ac)$ \(\Rightarrow\) $ab + bc +ac = -1 $
Ta có: $(a^2+b^2+c^2) = 2$

\(\Leftrightarrow\) $(a^2+b^2+c^2)^2 = 4$

\(\Leftrightarrow\)$a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2 = 4$

\(\Leftrightarrow\) $a^4+b^4+c^4 + 2(a^b^2+b^2c^2+a^2c^2) = 4$ (1)
Do $2(ab+bc+ac)^2 = 2(a^2b^2+b^2c^2+a^2c^2 + 2a^2bc+2ab^2c+2abc^2)$ (2)
Từ (1)(2) => $a^4+b^4+c^4+2(ab+bc+ac)^2 - 4abc(a+b+c) = 4$(3)
Thay $(ab+bc+ac) = -1$ và $a+b+c = 0$ (4)
Từ (3)(4) => $a^4 + b^4 + c^4 +2(-1)^2 -4abc.(0) = 4 $
<=> $a^4 + b^4 + c^4 + 2 = 4 => a^4 + b^4 + c^4 = 2 $

a+b+c=0

=>(a+b+c)²=0

=>a²+b²+c²+2(ab+bc+ca)=0

Do a²+b²+c²=1

=>2(ab+bc+ca)=-1

=>ab+bc+ca=-0,5

=>(ab+bc+ca)²=0,25

=>a²b²+b²c²+c²a²+2abc(a+b+c)=0,25

=>a²b²+b²c²+c²a²=0,25(do a+b+c=0)

Từ a²+b²+c²=1

=>(a²+b²+c²)²=1

=>a⁴+b⁴+c⁴+2(a²b²+b²c²+c²a²)=1

=>a⁴+b⁴+c⁴+2.0,25=1

=>a⁴+b⁴+c⁴+0,5=1

=>a⁴+b⁴+c⁴=0,5