ta co : 17⁵ + 24⁴ - 13²¹

=17⁴.17 + ...6 - 13^4.5.13

= ...7 + ...6 - ...3

=...0

vi 17⁵ + 24⁴ - 13²¹ co chu so tan cung bang 0 chia het cho 10 nen 17⁵+24⁴+13²¹ chia het cho 10

minh cha loi cau nay giup ban de xin loi ban cau truoc 

Ta có

 \(17^{25}=\left(17^2\right)^{12}.17=\left(\overline{.....9}\right)^{12}.17=\overline{......1}.17=\overline{.......7}\)

\(24^4=\overline{......6}\)

\(13^{21}=\left(13^2\right)^{10}.13=\left(\overline{.....9}\right)^{10}.13=\overline{.....1}.13=\overline{.....3}\)

\(\Rightarrow17^{25}+24^4-13^{21}=\overline{......7}+\overline{.....6}-\overline{.......3}=\overline{.......0}⋮10\)

\(\Rightarrow17^{25}+24^4-13^{21}⋮10\) (đpcm)

17²⁵=17.17²⁴=17.(17⁴)⁶=17.(...1)⁶ => Có tận cùng là 7

24⁴=(24²)²=(...6)² => Có tận cùng là 6

13²¹=13.13²⁰=13.(13⁴)⁵=13(...1)⁵ => Có tận cùng là 3

=> 17²⁵+24⁴-13²¹ =( ....7)+(....6)-(...3)=...0 => Có tận cùng là 0

=> 17²⁵+24⁴-13²¹ chia hết cho 10.

Bài 1:

CM: A = 17^5 + 24^4 + -13^21 chia hết cho 10

A = (17\(^4\)).17 + (24\(^2\))\(^2\) - (13\(^4\))\(^5\).13

A = \(\overline{..1}^{}.17\) + \(\overline{..6}^2\) - \(\overline{..1}^5.13\)

A = \(\overline{..7}\) + \(\overline{..6}\) - \(\overline{..3}\)

A = \(\overline{..3}\) - \(\overline{..3}\)

A = \(\overline{..0}\)

A chia hết cho 10 (đpcm)

BAI 2:chung minh rang voi moi so tu nhien n :

a) 7^4n -1 chia hết cho 5

A = 7\(^{4n}\)- 1

A = (7\(^4\))\(^{n}\) - 1

A = \(\overline{..1}\)\(^{n}\) - 1

A = \(\overline{..1}\) - 1

A= \(\overline{..0}\)

Vậy A chia hết cho 5

a)5\(^5\)-5\(^4\)+5\(^3\)=5\(^3\)x5\(^2\)-5\(^3\)x5\(^1\)+5\(^3\)x1=\(5^3\)x(\(5^2-5^1+1\))=\(5^3\)x121

BAI 1 :

CHO 3a + 2b chia het cho 17 ( a , b thuoc N ) . CHUNG MINH RANG : 10a + b chia het cho 17

Giải:

(3a + 2b) ⋮ 17

9(3a + 2b) ⋮ 17

(27a + 18b) ⋮ 17

(17a + 17b + 10a + b) ⋮ 17

(10a + b) ⋮ 17(đpcm)