Question

Cho a+b+c=0 va a^2+b^2+c^2=1     

Tinh gia tri cua A= a^4+b^4+c^4=

Answer 1

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

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