Question

[đỗ Thành Long]

1 câu nữa

Chứng minh

a) \(2010^{100}+2010^{99}⋮2011\)

b) \(3^{1994}+3^{1993}-3^{1992}⋮11\)

c) \(4^{13}+32^5-8^8⋮5\)

giúp nhé

xong trước 1 like \(☺\)

Answer 1

[๖Fly༉Donutღღ]

a)   \(2010^{100}\)+   \(2010^{99}\)

=   \(2010^{99}\)\(\left(2010+1\right)\)

=   \(2010^{99}\).   \(2011\)chia hết cho 2011

Vậy ...................................

b)   \(3^{1994}\)+   \(3^{1993}\)-   \(3^{1992}\)

=   \(3^{1992}\)\(\left(3^2+3-1\right)\)

=   \(3^{1992}\).   \(11\)

Vậy .......................

c)   \(4^{13}\)+   \(32^5\)-   \(8^8\)

=   \(\left(2^2\right)^{13}\)+   \(\left(2^5\right)^5\)-   \(\left(2^3\right)^8\)

=   \(2^{26}\)-   \(2^{25}\)-   \(2^{24}\)

=   \(2^{24}\).   \(\left(2^2+2-1\right)\)

=    \(2^{24}\). \(5\)

Vậy .......................

Answer 2

[Tiến Dũng Trương]

3 cau 3 nhe

a)

\(=2010^{99}\left(2010+1\right)\)

\(=2010^{99}.2011\) 

cung thay chia het ro nhi

b)

\(=3^{1992}\left(3^2+3-1\right)\)

\(=3^{1992}.11\)

cung thay chia het ro nhi

c)

\(=\left(2^2\right)^{13}+\left(2^5\right)^5-\left(2^3\right)^8\)

\(=2^{26}+2^{25}-2^{24}\)

\(=2^{24}\left(2^2+2-1\right)\)

\(=2^{24}.5\)

cung thay chia het ro nhi

cho 3 nhe