Question

[Nguyen Thi Dan Ha]

Chứng minh A=2019+2019^2+2019^3+2019^4+2019^5+2019^6 chia hết cho 2.

Answer 1

[╰Nguyễn Trí Nghĩa (team toán học)ッ...]

+)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

Answer 2

[Emma]

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) \(⋮2\rightarrowĐPCM\)

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