cho a^2 (b+c) +b^2(c+a) +c^2(a+b) +2abc = 0 . Cmr : a^2019 + b^2019 +c^2019 = (a+b+c)^2019

0 bình luận về “cho a^2 (b+c) +b^2(c+a) +c^2(a+b) +2abc = 0 . Cmr : a^2019 + b^2019 +c^2019 = (a+b+c)^2019”

1. Giải thích các bước giải:

   Ta có:

   $a^2(b+c)+b^2(c+a)+c^2(a+b)+2abc=0$

   $\to a^2b+a^2c+b^2c+b^2a+c^2(a+b)+2abc=0$

   $\to ( a^2b+b^2a)+(a^2c+b^2c+2abc)+c^2(a+b)=0$

   $\to ab(a+b)+c(a+b)^2+c^2(a+b)=0$

   $\to (a+b)(ab+c(a+b)+c^2)=0$

   $\to (a+b)(ab+ca+cb+c^2)=0$

   $\to (a+b)(a+c)(b+c)=0$

   Không mất tính tổng quát giả sử $a+b=0$

   $\to a=-b$

   $\to a^{2019}+b^{2019}+c^{2019}=(-b)^{2019}+b^{2019}+c^{2019}$

   $\to a^{2019}+b^{2019}+c^{2019}=-b^{2019}+b^{2019}+c^{2019}$

   $\to a^{2019}+b^{2019}+c^{2019}=0+c^{2019}$

   $\to a^{2019}+b^{2019}+c^{2019}=c^{2019}$

   $\to a^{2019}+b^{2019}+c^{2019}=(0+c)^{2019}$

   $\to a^{2019}+b^{2019}+c^{2019}=(a+b+c)^{2019}$

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