Question

[Bùi Béo]

cho a+b+c=0,a²+b²+c²=1.Tinh gia tri bieu thuc: A=a⁴+b⁴+c⁴

Answer 1

[Xyz OLM]

Ta có a + b + c = 0 

<=> (a + b + c)² = 0

<=> a² + b² + c² + 2(ab + bc + ca) = 0 

<=> ab + bc + ca = \(-\frac{1}{2}\)

=> \(\left(ab+bc+ca\right)^2=\frac{1}{4}\)

<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2ab^2c+2a^2bc+2abc^2=\frac{1}{4}\)

<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=\frac{1}{4}\)

<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\frac{1}{4}\)

Lại có a² + b² + c² = 1

=> (a² + b² + c²)² = 1

<= > a⁴ + b⁴ + c⁴ + 2[(ab)² + (bc)² + (ca)²] = 1 

<=> \(a^4+b^4+c^4+2.\frac{1}{4}=1\)

<=> \(a^4+b^4+c^4=\frac{1}{2}\)

Answer 2

[l҉o҉n҉g҉ d҉z҉]

Từ a + b + c = 0 => ( a + b + c )² = 0 <=> a² + b² + c² + 2ab + 2bc + 2ca = 0

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

<=> a²b² + b²c² + c²a² + 2ab²c + 2bc²a + 2a²bc = 1/4

<=> a²b² + b²c² + c²a² + 2abc( a + b + c ) = 1/4

<=> a²b² + b²c² + c²a² = 1/4 ( vì a + b + c = 0 )

Từ a² + b² + c² = 1 => ( a² + b² + c² )² = 1 <=> a⁴ + b⁴ + c⁴ + 2a²b² + 2b²c² + 2c²a² = 1

<=> a⁴ + b⁴ + c⁴ + 2( a²b² + b²c² + c²a² ) = 1 

<=> a⁴ + b⁴ + c⁴ + 1/2 = 1 <=> a⁴ + b⁴ + c⁴ = 1/2

Vậy A = 1/2