Ta có a² + b² + c² = 14

=> (a² + b² + c²)² = 196

=> a⁴ + b⁴ + c⁴ + 2a²b² + 2b²c² + 2c²a² = 196

=> a⁴ + b⁴ + c⁴ + 2(a²b² + b²c² + c²a²) = 196

Lại có a + b + c = 0

=> (a + b + c)² = 0

=> a² + b² + c² + 2ab + 2bc + 2ca = 0

=> 2(ab + bc + ca) = -14

=> ab + bc + ca = -7

=> (ab + bc + ca)² = 49

=> a²b² + b²c² + c²a² + 2ab²c + 2a²bc + 2abc² = 49

=> a²b² + b²c² + c²a² + 2abc(a + b + c) = 49

=> a²b² + b²c² + c²a² = 49

Khi đó a⁴ + b⁴ + c⁴ + 2(a²b² + b²c² + c²a²) = 196

<=> a⁴ + b⁴ + c⁴ + 2.49 = 196

=>  a⁴ + b⁴ + c⁴ + 98 = 196

=> a⁴ + b⁴ + c⁴ = 98

Vậy N = 98

Ta có a + b + c = 0

=> a + b = -c

=> (a + b)² = (-c)²

=> a² + b² + 2ab = c²

=> a² + b² - c² = -2ab

=> (a² + b² - c²)² = (-2ab)²

=> a⁴ + b⁴ + c⁴ + 2a²b² - 2a²c² - 2b²c² = 4a²b²

=> a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c² + 2a²c²

Khi đó a² + b² + c² = 14

<=> (a² + b² + c²)² = 14²

=> a⁴ + b⁴ + c⁴ + 2a²b² + 2b²c² + 2a²c² = 196

=> a⁴ + b⁴ + c⁴ + a⁴ + b⁴ + c⁴ = 196 (Vì a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c² + 2a²c²)

=> 2(a⁴ + b⁴ + c⁴) = 196

=> a⁴ + b⁴ + c⁴ = 98

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 ...

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

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

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

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

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

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

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

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

bạn ơi nó phải bằng 2 chứ

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

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