Question

cho a+b+c=1 và a^3+b^3+c^3=1

CMR a^2005+b^2005+c^2005=1

Answer 1

Ta có : \(a+b+c=1\)

\(\Leftrightarrow\left(a+b+c\right)^3=1\)

\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)=1\)

\(\Leftrightarrow1+3\left(a+b\right)\left(b+c\right)\left(a+c\right)=1\)

\(\Leftrightarrow3\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\b+c=0\\a+c=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)

Với \(a=-b\) , \(a+b+c=1\)

\(\Rightarrow c=1\)

\(\Rightarrow a^{2005}+b^{2005}+c^{2005}=\left(-b\right)^{2005}+b^{2005}+c^{2005}=c^{2005}=1^{2005}=1\left(1\right)\)

Với \(b=-c\) , \(a+b+c=1\)

\(\Rightarrow a=1\) CMTT , ta được :

\(a^{2005}+b^{2005}+c^{2005}=1\left(2\right)\)

Với \(c=-a\) , \(a+b+c=1\)

\(\Rightarrow b=1\) CMTT , ta được :

\(a^{2005}+b^{2005}+c^{2005}=1\left(3\right)\)

Từ ( 1 ) ; ( 2 ) ; ( 3 )

\(\Rightarrow a^{2005}+b^{2005}+c^{2005}=1\left(đpcm\right)\)

P/s : Làm linh tinh , ko chắc :D

Answer 2

Link c/m : \(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

https://hoc24.vn/hoi-dap/question/668753.html