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

<=> 2+2(ab+bc+ca)=0 => ab+bc+ca=-1

(ab+bc+ca)^2=(ab)^2+(bc)^2+(ca)^2+2ab^2c+2abc^2+2a^2bc=(ab)^2+(bc)^2+(ca)^2+2abc(a+b+c)

=> (ab)^2+(bc)^2+(ca)^2 = (-1)^2 = 1

(a^2+b^2+c^2)^2 = a^4+b^4+c^4+2[(ab)^2+(bc)^2+(ca)^2] = a^4+b^4+c^4 + 2

<=>4=a^4+b^4+c^4+2 => a^4+b^4+c^4 = 2

Bạn kiểm tra lại có sai chỗ nào không nhé

1/2 nhá

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

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)

lại nhầm lần này đúng

(a+b+c)²=a²+b²+c²+2ac+2bc+2ab

=>0²=2+2(ac+bc+ab)

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

=>(-1)²=a²b²+b²c²+a²c²+2a²bc+2b²ac+2c²ab

(-1)²=a²b²+b²c²+a²c²+2abc(a+b+c)

=>1=a²b²+b²c²+a²c²+2abc.0

=>a²b²+b²c²+a²c²=1

(a²+b²+c²)²=a⁴+b⁴+c⁴+2a²b²+2b²c²+2a²c²

(a²+b²+c²)²=a⁴+b⁴+c⁴+2(a²b²+b²c²+a²c²)

2²=a⁴+b⁴+c⁴+2.1

4=a⁴+b⁴+c⁴+2

=>a⁴+b⁴+c⁴=2

trieu dang   làm sai đoạn cuối rồi

(a+b+c)²=a²+b²+c²+2ac+2bc+2ab

=>0²=2+2(ac+bc+ab)

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

=>(-1)²=a²b²+b²c²+a²c²+2a²bc+2b²ac+2c²ab

(-1)²=a²b²+b²c²+a²c²+2abc(a+b+c)

=>1=a²b²+b²c²+a²c²+2abc.0

=>a²b²+b²c²+a²c²=1

(a²+b²+c²)²=a⁴+b⁴+c⁴+2a²b²+2b²c²+2a²c²

(a²+b²+c²)²=a⁴+b⁴+c⁴+2(a²b²+b²c²+a²c²)

2²=a⁴+b⁴+c⁴+2.1

4=a⁴+b⁴+c⁴+2

=>a⁴+b⁴+c⁴=2

\(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

\(\Leftrightarrow2+2\left(ab+bc+ca\right)=0\Leftrightarrow ab+bc+ca=-1\Rightarrow\left(ab+bc+ca\right)^2=1\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc=1\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=1\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc.0=1\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=-1\)

Xét \(a^2+b^2+c^2=2\Rightarrow\left(a^2+b^2+c^2\right)^2=4\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(-1\right)=4\Leftrightarrow a^4+b^4+c^4=6\)

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