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AH
Akai Haruma
Giáo viên
3 tháng 11 2023

Lời giải:

$A=(5+5^2)+(5^3+5^4)+...+(5^{29}+5^{30})$

$=(5+5^2)+5^2(5+5^2)+....+5^{28}(5+5^2)$

$=(5+5^2)(1+5^2+....+5^{28})=30(1+5^2+...+5^{28})\vdots 30$

em cảm ơn cô ạ

 

27 tháng 12 2017

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28 tháng 12 2017

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8 tháng 8 2015

Ta xét: (a^5 - a) + (b^5 - b) + (c^5 - c)

Ta có: a^5 - a = a(a^4 - 1) = a(a² - 1)(a² + 1) = a(a - 1)(a + 1)(a² + 1) 
= a(a - 1)(a + 1)(a² - 4 + 5) 
= a(a - 1)(a + 1)[ (a² - 4) + 5) ] 
= a(a - 1)(a + 1)(a² - 4) + 5a(a - 1)(a + 1) 
= a(a - 1)(a + 1)(a - 2)(a + 2) + 5a(a - 1)(a + 1) 
= (a - 2)(a - 1)a(a + 1)(a + 2) + 5a(a - 1)(a + 1) 

Do (a - 2)(a - 1)a(a + 1)(a + 2) là tích của 5 số nguyên liên tiếp => (a - 2)(a - 1)a(a + 1)(a + 2) chia hết cho 2, 3, 5 và 5a(a - 1)(a + 1) chia hết cho 5 và 2, 3 hay chia hết cho 2*3*5=30 

=> (a - 2)(a - 1)a(a + 1)(a + 2) + 5a(a - 1)(a + 1) chia hết cho 30. 

=> a^5 - a chia hết cho 30 

=> (a^5 -a) + (b^5 -b) + (c^5 -c) = (a^5+b^5+c^5) -(a+b+c) chia hết cho 30 (*) 

Do (a+b+c) chia hết cho 30 

(*) => (a^5+b^5+c^5) chia hết cho 30

Đó là câu trả lời đúng.hihi :) 

Ta xét (a^5 -a) + (b^5 -b) + (c^5 -c) 

Ta có: a^5 - a = a(a^4 - 1) = a(a² - 1)(a² + 1) = a(a - 1)(a + 1)(a² + 1) 
= a(a - 1)(a + 1)(a² - 4 + 5) 
= a(a - 1)(a + 1)[ (a² - 4) + 5) ] 
= a(a - 1)(a + 1)(a² - 4) + 5a(a - 1)(a + 1) 
= a(a - 1)(a + 1)(a - 2)(a + 2) + 5a(a - 1)(a + 1) 
= (a - 2)(a - 1)a(a + 1)(a + 2) + 5a(a - 1)(a + 1) 

Do (a - 2)(a - 1)a(a + 1)(a + 2) là tích của 5 số nguyên liên tiếp => (a - 2)(a - 1)a(a + 1)(a + 2) chia hết cho 2, 3, 5 và 5a(a - 1)(a + 1) chia hết cho 5 và 2, 3 hay chia hết cho 2*3*5=30 

=> (a - 2)(a - 1)a(a + 1)(a + 2) + 5a(a - 1)(a + 1) chia hết cho 30. 

=> a^5 - a chia hết cho 30 

=> (a^5 -a) + (b^5 -b) + (c^5 -c) = (a^5+b^5+c^5) -(a+b+c) chia hết cho 30 (*) 

Do (a+b+c) chia hết cho 30 

(*) => (a^5+b^5+c^5) chia hết cho 30

14 tháng 4 2018

a) 40232 ;     b) 1245 ;     c) 52110 ;     d) 1245 ;     e) 52110

4 tháng 8 2019

Ta thấy : \(a^5-a=a\left(a^4-1\right)=a\left(a^2-1\right)\left(a^2+1\right).\)

\(=a\left(a-1\right)\left(a+1\right)\left(a^2-4+5\right)\)

\(=a\left(a-1\right)\left(a+1\right)\left(a^2-4\right)+5a\left(a-1\right)\left(a+1\right)\)

\(=\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)+5a\left(a-1\right)\left(a+1\right)\)

Ta có :\(\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)\)là tích 5 số tự nhiên liên tiếp :

\(\Rightarrow\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)\)\(⋮\)\(5\)và cũng \(⋮\)\(6\)( cũng là 3 số tự nhiên liên tiếp )

\(\Rightarrow\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)\)\(⋮\)\(30\)\(\left(1\right)\)

Ta lại có : \(5\)\(⋮\)\(5\)và \(\left(a-1\right)a\left(a+1\right)\)\(⋮\)\(6\)

\(\Rightarrow5a\left(a-1\right)\left(a+1\right)\)\(⋮\)\(30\)\(\left(2\right)\)

Từ ( 1 ) và ( 2 ) \(\Rightarrow\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)+5a\left(a-1\right)\left(a+1\right)\)\(⋮\)\(30\)

Hay \(a^5-a\)\(⋮\)\(30\)

Tương tự \(b^5-b\)và \(c^5-c\)cũng chia hết cho 30 

\(\Rightarrow a^5+b^5+c^5-\left(a+b+c\right)\)\(⋮\)\(30\)

Mà \(a+b+c\)\(⋮\)\(30\)

\(\Rightarrow a^5+b^5+c^5\)\(⋮\)\(30\)\(\left(đpcm\right)\)

25 tháng 10 2017

28 tháng 12 2019

a, Số nào chia hết cho 2 mà không chia hết cho 5: 422

b, Số nào chia hết cho 5 mà không chia hết cho 2: 105

c, Số nào chia hết cho cả 2 và 5: 6760

d, Số nào không chia hết cho cả 2 và 5: 3071

27 tháng 9 2017

a, Số chia hết cho 2 mà không chia hết cho 5 là: 844

b, Số nào chia hết cho 5 mà không chia hết cho 2 là: 105

c, Số nào chia hết cho cả 2 và 5 là: 6740

d, Số nào không chia hết cho cả 2 và 5 là: 3071