a, \(a^3+b^3+c^3=3abc\)
⇔\(a^3+b^3+c^3-3abc=0\)
⇔\(\left(a+b\right)^3+c^3-3abc-3a^2b-3ab^2=0\)
⇔\(\left(a+b+c\right)\left(\left(a+b\right)^2-\left(a+b\right)c+c^2\right)-3ab\left(a+b+c\right)=0\)
⇔\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc-3ab\right)=0\)
⇔\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
⇒\(a^2+b^2+c^2-ab-bc-ac=0\left(a+b+c\ne0\right)\)
⇔\(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
⇔\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
⇔\(a=b=c\)
⇒\(=\frac{a^{2016}+a^{2016}+a^{2016}}{\left(a+a+a\right)^{2016}}=\frac{3a^{2016}}{3^{2016}\cdot a^{2016}}=\frac{1}{3^{2015}}\)
b/ \(n^2+4n+2013=k^2\) (\(k\in N\))
\(\Leftrightarrow\left(n+2\right)^2+2009=k^2\)
\(\Leftrightarrow k^2-\left(n+2\right)^2=2009\)
\(\Leftrightarrow\left(k-n-2\right)\left(k+n+2\right)=2009=1.2009=7.287=41.49\)
Do \(k-n-2< k+n+2\) nên ta chỉ cần xét 3 trường hợp:
\(\left\{{}\begin{matrix}k-n-2=1\\k+n+2=2009\end{matrix}\right.\) \(\Rightarrow2n+4=2008\Rightarrow n=1002\)
\(\left\{{}\begin{matrix}k-n-2=7\\k+n+2=287\end{matrix}\right.\) \(\Rightarrow n=138\)
\(\left\{{}\begin{matrix}k-n-2=41\\k+n+2=49\end{matrix}\right.\) \(\Rightarrow n=2\)
Vậy \(n=\left\{2;138;1002\right\}\)
là: M = ...........