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Ta có: \(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)
\(\Rightarrow5A=5+5^2+5^3+...+5^{50}+5^{51}\)
\(\Rightarrow5A-A=\left(5+5^2+5^3+...+5^{50}+5^{51}\right)-\left(1+5+5^2+...+5^{49}+5^{50}\right)\)
\(\Rightarrow4A=5^{51}-1\)
\(\Rightarrow A=\frac{5^{51}-1}{4}\)
Vậy \(A=\frac{5^{51}-1}{4}\)
5A=5+5^2+5^3+5^4+.....+5^50+5^51
A=1+5+5^2+5^3+....+5^49+5^50
=>5A-A=(5+5^2+5^3+5^4+...+5^50+5^51)-(1+5+5^2+5^3+....+5^49+5^50)
=5^51-1
=>A=\(\frac{5^{51}-1}{4}\)
tick nhe1
a)Đặt \(A=7^6+7^5-7^4\)
\(A=7^4\left(7^2+7-1\right)\)
\(A=7^4\cdot55⋮55\left(đpcm\right)\)
b)\(A=1+5+5^2+5^3+...+5^{50}\)
\(5A=5+5^2+5^3+5^4+...+5^{51}\)
\(5A-A=\left(5+5^2+5^3+5^4+...+5^{51}\right)-\left(1+5+5^2+5^3+...+5^{50}\right)\)
\(4A=5^{51}-1\)
\(A=\frac{5^{51}-1}{4}\)
a)
Ta có :
\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55\)
=> Chia hết cho 5
b)
Ta có :
\(A=1+5+5^2+....+5^{50}\)
\(5A=5+5^2+....+5^{51}\)
=> 5A - A = \(\left(5+5^2+....+5^{51}\right)\)\(-\left(1+5+....+5^{50}\right)\)
\(\Rightarrow4A=5^{51}-1\)
\(\Rightarrow A=\frac{5^{51}-1}{4}\)
A =1+5+52+53+...+549+550
5A = 5 + 52 + 53 + 54 + ...+ 550+ 551
5A -A = 5 + 52 + 53 + 54 + ...+ 550+ 551 - 1- 5 - 52-53- ... - 549-550
4A= 551 -1
A= 551 -1 / 4
5A=5+52+53+...+550+551
5A-A=551-1
A=551-1:4
tick mk nha cái kia sai rôi
a) 76 + 75 - 74 = 74 ( 72 + 7 - 1) = 74 . 55\(⋮\)55
b) A = 1 + 5 + 52 + ... + 550
5A = 5 + 52 + 53 + ... + 551
5A - A = ( 5 + 52 + 53 + ... + 551) - ( 1 + 5 + 52 + ... + 550)
4A = 551 - 1
A = \(\frac{5^{51}-1}{4}\)
Ta có :A = 1 + 5 + 52 + 53 + .... + 549 + 550
=> 5A = 5 + 52 + 53 + .... + 550 + 551
=> 5A - A = 551 - 1
=> 4A = 551 - 1
=> A = \(\frac{5^{51}-1}{4}\)
Ta có:
A=1+5+52+53+...+550
=> 5A=5+52+53+...+550+551
=>5A-A=(5+52+53+...+550+551)-(1+5+52+53+...+550)
=> 4A=551-1
=>A=\(\dfrac{5^{51}-1}{4}\)
Ta có:
A = 1+ 5 + 52 + 53 + ......... + 549 + 550
=> 5A = 5 + 52 + 53 + 54 +.......+ 549 + 550
Do đó: 5A - A = 551 - 1
Vậy A = \(\frac{5^{51}-1}{4}\)
5A = 5+5^2+5^3+....+5^51
5A - A = (5-5)+(5^2-5^2)+....+(5^50-5^50) + 5^51-1
4A = 5^51 - 1
\(\Rightarrow A=\frac{5^{51}-1}{4}\)
Câu hỏi tương tự có đó Hermione Granger