Ta cm : n^5-n có chữ số tận cùng = 0 

Ta có : \(n^5-n=n\left(n^4-1\right)=n\left(n^2-1\right)\left(n^2+1\right)\\ n⋮2\Rightarrow A⋮2\\ nko⋮2\Rightarrow n^2-1;n^2+1⋮2\Rightarrow A⋮2\)

\(n⋮3\Rightarrow A⋮3\\ nko⋮3\\ \Rightarrow n^2chia3duw1\\ \Rightarrow n^2-1⋮3\\ \Rightarrow A⋮3\)

\(n⋮5\Rightarrow A⋮5\\ nko⋮5\Rightarrow n^2chia5du1;4\\ n^2:5du1\\ \Rightarrow n^2-1⋮5\\ \Rightarrow A⋮5\\ n^2:5du4\\ \Rightarrow n^2+1⋮5\\ \Rightarrow A⋮5\)

(2;3;5) ntoCN từng đôi => n^5-n chia hết cho 30 

=> n^5-n có t/c = 0 

=> đpcm 

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\(\left(x-1\right)\left(x+5\right)=x^2+4x-5\)(1)

\(x\left(x+4\right)=x^2+4x\)(2)

Lấy (1) nhân (2) \(\Leftrightarrow y.\left(y-5\right)=84\Leftrightarrow y^2-5y+\left(\frac{5}{2}\right)^2=84+\frac{25}{4}=\left(\frac{19}{2}\right)^2\)

\(\orbr{\begin{cases}y=\frac{5-19}{2}=-7\left(loai\right)\\y=\frac{5+19}{2}=12\end{cases}}\) 

\(x^2+4x=12\Leftrightarrow\left(x+2\right)^2=16\Rightarrow\orbr{\begin{cases}x=-6\\x=2\end{cases}}\)

\(A=2012-\left(2x^2+5y^2-2xy-4x-4y\right)\\ \)

Hệ số lẻ quá:

B=2A đặt 2x=z

\(B=m-\left(z^2+10y^2-2yz-4z-8y\right)\)

\(B=m-\left[\left(z-y-2\right)^2+9y^2-12y-4\right]\)

\(B=m-\left[\left(z-y-2\right)^2+\left(t-2\right)^2-4-4\right]\)

\(B=\left(m-8\right)-\left(z-y-2\right)^2-\left(t-2\right)^2\)

\(A_{min}=\frac{2.2012-8}{2}=2008\)đạt tại  \(\orbr{\begin{cases}t-2=0=>y=\frac{2}{3}\\z-y-2=0\Rightarrow x=\frac{4}{3}\end{cases}}\)

ghép BP là ra thôi

\(\hept{\begin{cases}x^2=8x+y\\y^2=8y+x\end{cases}\Leftrightarrow\left(x-y\right)\left(x+y\right)=7\left(x-y\right)vi.x\ne}y\Leftrightarrow x+y=7\)

\(\Leftrightarrow x^2+y^2=8\left(x+y\right)+\left(y+x\right)\Rightarrow DS=9\cdot7=63\)