Question

cho a+b+c=0 cm

a a^4+b^4+c^4=2(a^2.b^2+b^2.c^2+c^2.a^2)

Answer 1

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Answer 2

Từ a+b+c=0 có b+c =-a 
Suy ra (b+c)^2 = (-a)^2 hay b^2 + c^2 +2bc = a^2 
hay b^2 + c^2 -a^2 = -2bc 

Suy ra (b^2 + c^2 - a^2)^2 = (-2bc)^2 
<=> b^4 + c^4 + a^4 +2b^2.c^2 - 2a^2.b^2 - 2a^2.c^2 = 4b^2.c^2 
<=> a^4 + b^4 + c^4 = 2a^2.b^2 + 2b^2.c^2 + 2c^2.a^2 
<=> 2(a^4 + b^4 + c^4) =a^4 + b^4 + c^4 + 2a^2.b^2 + 2b^2.c^2 + 2c^2.a^2 
<=> 2(a^4 + b^4 + c^4 ) =(a^2 + b^2 + c^2): Đpcm

Answer 3

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

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

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

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

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)\) (vì a+b+c=0)

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