+)Ta có:\(A=2019+2019^2+2019^3+2019^4+2019^5+2019^6\)

\(\Rightarrow A=\left(2019+2019^2\right)+\left(2019^3+2019^4\right)+\left(2019^5+2019^6\right)\)

\(\Rightarrow A=\left(2019+2019^2\right)+2019^2.\left(2019+2019^2\right)+2019^4.\left(2019+2019^2\right)\)

+)Ta lại có:2019² tận cùng là 1

=>2019+2019² tân cùng là 9+1=10

=>2019+2019²\(⋮2\)

\(\Rightarrow\left(2019+2019^2\right)⋮2;2019^2.\left(2019+2019^2\right)⋮2;2019^4.\left(2019+2019^2\right)⋮2\)

\(\Rightarrow A⋮2\)

Vậy \(A⋮2\left(ĐPCM\right)\)

Chúc bn học tốt

A = 2019 + 2019² + 2019³ + 2019⁴ + 2019⁵ + 2019⁶

A = ( 2019 + 2019² ) + ( 2019³ + 2019⁴) + ( 2019⁵ + 2019⁶)

A = 1 . ( 2019 + 2019² ) + 2019³ . (2019 + 2019² ) + 2019⁵ . ( 2019 + 2019² )

A = 1 . 4 078 380   + 2019³ . 4 078 380 + 2019⁵ . 4 078 380

A = 4 078 380 . ( 1 + 2019³ + 2019⁵) \(⋮2\rightarrowĐPCM\)

# HOK TỐT #

        \(A=2019+2019^2+2019^3+2019^4+2019^5+2019^6\)

<=> \(A=\left(2019+2019^2\right)+\left(2019^3+2019^4\right)+\left(2019^5+2019^6\right)\)

<=>\(A=2019.\left(1+2019\right)+2019^3.\left(1+2019\right)+2019^5\left(1+2019\right)\)

<=>\(A=2019.2020+2019^3.2020+2019^5.2020\)

<=>\(A=2020.\left(2019+2019^3+2019^5\right)\)

<=>\(A=2.1010\left(2019+2019^3+2019^5\right)⋮2\)=> \(A⋮2\)

Vậy .....

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\(2019^{2019}-1=2019^{2019}-1^{2019}⋮2019-1=2018⋮2018\)

\(\frac{3x-3}{6}=\frac{2y+10}{10}=\frac{5z-10}{15}=\frac{3x+2y-5z+17}{1}=\frac{3x+2y-5z+16+1}{1}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x-1}{2}=1\\\frac{y+5}{5}=1\\\frac{z-2}{3}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=0\\z=5\end{matrix}\right.\)

\(\Rightarrow P=3^{2019}+5^{2019}\)

Ta có \(3\equiv-1\left(mod4\right)\Rightarrow3^{2019}\equiv-1\left(mod4\right)\)

\(5\equiv1\left(mod4\right)\Rightarrow5^{2019}\equiv1\left(mod4\right)\)

\(\Rightarrow P\equiv\left(-1+1\right)\left(mod4\right)\Rightarrow P\equiv0\left(mod4\right)\Rightarrow P⋮4\)