thêm đk \(a\in Z\)
\(M=a^3-a+6a\)
\(\Rightarrow M=a\left(a^2-1\right)+6a\)
\(\Rightarrow M=\left(a-1\right)a\left(a+1\right)+6a\)
+ \(\left(a-1\right)a\left(a+1\right)\) là tích 3 số nguyên liên tiếp
\(\Rightarrow\left\{{}\begin{matrix}\left(a-1\right)a\left(a+1\right)⋮2\\\left(a-1\right)a\left(a+1\right)⋮3\end{matrix}\right.\)
\(\Rightarrow\left(a-1\right)a\left(a+1\right)⋮6\)
\(\Rightarrow\left(a-1\right)a\left(a+1\right)+6a⋮6\)
\(\Rightarrow M⋮6\)
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