\(a+b+c=0\)

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

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

\(\Leftrightarrow2ab+2bc+2ca=-14\)

\(\Leftrightarrow ab+bc+ca=-7\)

\(\Rightarrow\left(ab+bc+ca\right)^2=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=49\).

\(a^2+b^2+c^2=14\)

\(\Rightarrow\left(a^2+b^2+c^2\right)^2=14^2=196\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=196\)

\(\Leftrightarrow a^4+b^4+c^4+2.49=196\)

\(\Leftrightarrow a^4+b^4+c^4=98\)

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

\(\Leftrightarrow14+2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=-7\)

Suy ra : \(\left(ab+bc+ac\right)^2=49\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)

\(a^2+b^2+c^2=14\Leftrightarrow\left(a^2+b^2+c^2\right)^2=196\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)

\(\Leftrightarrow a^4+b^4+c^4+2.49=256\)  \(\Leftrightarrow a^4+b^4+c^4=98\)

Vậy ... 

Xét đẳng thức a² + b² + c² = 0, ta có :

\(a^2\ge0\)

\(b^2\ge0\)  => a² + b² + c² \(\ge0\)

\(c^2\ge0\)

Mà đề cho a² + b² + c² = 0

=> \(\hept{\begin{cases}a^2=0\\b^2=0\\c^2=0\end{cases}\Rightarrow\hept{\begin{cases}a=0\\b=0\\c=0\end{cases}}}\)

Đồng thời nó cũng thõa mãn điều kiện a + b + c = 0

Ta có :

a⁴ + b⁴ + c⁴  = 0 + 0 + 0 = 0

đề nghị bn Kurosaki ko làm đc thì đừng giải tầm bậy nhé , người khác học theo cách giải của bn thì hậu quả thế nào,đã bao giờ mở mang đầu óc như vậy chưa ?

Ta có: a+b+c=0

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

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

\(\Leftrightarrow2\left(ab+bc+ac\right)=0-1=-1\)

hay \(ab+bc+ac=-\dfrac{1}{2}\)

\(\Leftrightarrow\left(ab+bc+ac\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2ab^2c+2abc^2+2a^2bc=\dfrac{1}{4}\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(b+c+a\right)=\dfrac{1}{4}\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=\dfrac{1}{4}\)

Ta có: \(M=a^4+b^4+c^4\)

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

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

\(\Leftrightarrow M=1^2-2\cdot\dfrac{1}{4}=1-\dfrac{1}{2}=\dfrac{1}{2}\)

Vậy: \(M=\dfrac{1}{2}\)

Ta có : \(a+b+c=0\)

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

\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)=1\) ( * )

\(\Rightarrow ab+bc+ac=-\dfrac{1}{2}\)

Lại có : \(\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\) ( suy ra từ * )

\(\Rightarrow a^4+b^4+c^4=2\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{2}\)

Vậy ...

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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