Ta có:

\(A=5^{2014}-5^{2013}+5^{2012}\)

\(A=5^{2011}\left(5^3-5^2+5\right)\)

\(A=5^{2011}\left(125-25+5\right)\)

\(A=5^{2011}.105\)

\(\Rightarrow A⋮105\)

=> ĐPCM.

Ta có: \(5^{2014}-5^{2013}+5^{2012}=5^{2011}\left(5^3-5^2+5\right)\)

\(=5^{2011}.105⋮105\)

\(\Rightarrow5^{2014}-5^{2013}+5^{2012}⋮105\left(đpcm\right)\)

Vậy...

ta có:

\(5^{2014}-5^{2013}+5^{2012}\)

\(=5^{2012}\left(5^2-5+1\right)\)

\(=5^{2012}\left(25-5+1\right)\)

\(=5^{2012}.21\)

ta thấy: \(5^{2012}.21⋮21\)

\(5^{2012}.21⋮5\)

\(\Rightarrow5^{2012}.21⋮21.5\)

\(\Rightarrow5^{2012}.21⋮105\)

\(\Leftrightarrow5^{2014}-5^{2013}+5^{2012}⋮105\left(đpcm\right)\)

5²⁰¹⁴-5²⁰¹³+5²⁰¹²

=5²⁰¹¹*5³-5²⁰¹¹*5²+5²⁰¹¹*5

=\(5^{2011}\cdot\left(5^3-5^2+5\right)\)

\(=5^{2011}\cdot105\)chia hết cho 105

\(5^{2014}-5^{2013}+5^{2012}=5^{2011}\left(5^3-5^2+5\right)\)

\(=5^{2011}.\left(125-25+5\right)=5^{2011}.105⋮105\)

thank bạn nha

Ta có :

\(3^{2014}+3^{2013}-3^{2012}\)

\(=3^{2012}\left(3^2+3-1\right)\)

\(=3^{2012}.11\)

\(\Rightarrow3^{2014}+3^{2013}-3^{2012}\)

\(\RightarrowĐPCM\)

\(A=5^{2014}-5^{2013}+5^{2012}=5^{2012}\left(5^2-5^1+5^0\right)=21.5^{2012}\\ \)

\(\hept{\begin{cases}105=21.5\\A=21.5^{2012}\end{cases}}\Rightarrow\frac{A}{105}=\frac{21.5^{2012}}{21.5}=5^{2011}\Rightarrow dpcm\)

5^2014-5^2013+5^2012=5^2012(5^2-5^1+1)

                                  =5^2012.21

                                  =5^2011.5.21

                                  =5^2011.105

Vậy 5^2014-5^2013+5^2012 chia hết cho 105

1) \(23^{401}+38^{202}-2^{433}=23^{4.100}.23+38^{4.50}.38^2-2^{4.108}.2^1=\left(..1\right).23+\left(..6\right).1444-\left(..6\right).2=\left(..3\right)+\left(..4\right)-\left(..2\right)=\left(..5\right)\)

làm các con kia tương tự nhé ^^

       \(5^{2014}-5^{2013}+5^{2012}\)

\(=5^{2011}.\left(5^3-5^2+5\right)\)

\(=5^{2011}.105\)\(⋮105\)

\(\Rightarrow5^{2014}-5^{2013}+5^{2012}⋮105\)\(\left(đpcm\right)\)