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.

       \(5^{2014}-5^{2013}+5^{2012}\)

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

\(=5^{2011}.105\)\(⋮105\)

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

Ta có : 5²⁰¹⁴ - 5²⁰¹³ + 5²⁰¹²

        = 5²⁰¹².(5² - 5 + 1)

        = 5²⁰¹².21

        = 5²⁰¹¹.5.21

        = 5²⁰¹¹.105 \(⋮\)105

=> 5²⁰¹⁴ - 5²⁰¹³ + 5²⁰¹² \(⋮\)105

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

= \(5^{2011}.\left(5^3-5^2+5\right)\)

= \(5^{2011}.105⋮105\)(ĐPCM)

5²⁰¹⁴ - 5²⁰¹³ + 5²⁰¹² = 5²⁰¹² . ( 5² - 5 + 1 ) = 5²⁰¹² . 21

5²⁰¹² chia hết cho 5, 21 chia hết cho 21 mà 5 với 21 là hai số nguyên tố cùng nhau nên 5²⁰¹² . 21 chia hết cho 105 

<33

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 5²⁰¹⁴ - 5²⁰¹³ + 5²⁰¹²

    = 5²⁰¹².(5² - 5 + 1)

    = 5²⁰¹².21

    = 5²⁰¹¹.5.21

    = 5²⁰¹¹.105 \(⋮\)105

=>  5²⁰¹⁴ - 5²⁰¹³ + 5²⁰¹² \(⋮\)105

Ta có :

5²⁰¹⁴ - 5²⁰¹³ + 5²⁰¹² 

= 5²⁰¹² . (5² - 5¹ + 1)

= 5²⁰¹² . (25 - 5 + 1)

= 5²⁰¹² . 21

= 5²⁰¹¹ . 5 . 21

= 5²⁰¹¹ . 105\(⋮\)105

=> 5²⁰¹⁴ - 5²⁰¹³ + 5^{2012 \(⋮\)}105 (đpcm)

~Study well~

#Zu

#)Giải :

\(5^{2014}-5^{2013}+5^{2012}=5^{2012}\left(5^2-5^6+1\right)=5^{2012}.21=5^{2011}.5.21=5^{2011}.105\)chia hết cho 105

\(\Rightarrowđpcm\)

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

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

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

 5 ^ 2014 - 5 ^2013 + 5 ^ 2012 = 5^2012 ( 5 ^ 2 - 5 6 1 + 1 )

=5 ^ 2012 . 21   =  5 ^ 2011 . 5 . 21   =  5 ^ 2011 . 105 

Vậy ................

\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)

=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)