Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : A = 20002016 + 20002017
= 20002016.(1 + 2000)
= 20002016.2001
< 20012016.2001
= 20012017 = B
=> A < B
Vậy A < B
B=20002017+2017 ,A=20002016+20002017
Mà 20002016>2017
=>A>B
ta có: \(A=\frac{1999^{1999}+1}{1999^{1998}+1}=\frac{1999.\left(1999^{1998}+1\right)-1998}{1999^{1998}+1}=\frac{1999.\left(1999^{1998}+1\right)}{1999^{1998}+1}-\frac{1998}{1999^{1998}+1}\)
\(=1999-\frac{1998}{1999^{1998}+1}\)
\(B=\frac{1999^{2000}+1}{1999^{1999}+1}=\frac{1999.\left(1999^{1999}+1\right)-1998}{1999^{1999}+1}=\frac{1999.\left(1999^{1999}+1\right)}{1999^{1999}+1}-\frac{1998}{1999^{1999}+1}\)
\(=1999-\frac{1998}{1999^{1999}+1}\)
mà \(\frac{1998}{1999^{1998}+1}>\frac{1998}{1999^{1999}+1}\Rightarrow1999-\frac{1998}{1999^{1998}+1}< 1999-\frac{1998}{1999^{1999}+1}\)
\(\Rightarrow A< B\)
ta có
175 = 17 . 174 = 17 . ( ...01) = ...7
244 = ...6
1321 = 13 . 1320 = 13 . (...1) = ...3
\(\Rightarrow\) 175 + 24 - 1321 = (...7) + (...6) - (...3) = ...0
vậy 175 +244 - 1321 có chữ số tận cùng bằng 0
A=4+4^2+4^3+...+4^50
A=(4+4^2)+(4^3+4^4)+...+(4^49+4^50)
A=(4+4^2)+4^2(4+4^2)+...+4^48(4+4^2)
A=4+4^2(4^2 +4^4+...+4^48)\(⋮\)10 (vì 4+4^2=20\(⋮\)10)
Vậy A\(⋮\)10
\(\frac{1999^{1999+1}}{1999^{2000+1}}=1-\frac{1}{1999^{2000+1}};\)\(\frac{1999^{1998+1}}{1999^{1999+1}}=1-\frac{1}{1999^{1999+1}}\)
Vì \(1-\frac{1}{1999^{2000+1}}< 1-\frac{1}{1999^{1999+1}}\)nên \(\frac{1999^{1999+1}}{1999^{2000+1}}>\frac{1999^{1998+1}}{1999^{1999+1}}\)
3:
a) Ta có:
44433 = 4443 . 11 = (4443)11 = 87 528 38411
33344 = 3334 . 11 = (3334)11 = 12 296 370 32111
Vì 87 528 38411 < 12 296 370 32111 nên 44433 < 33344
Vậy,...
A=1999+19992+....+1999204
A=(1999+19992)+....+(1999203+1999204)
A=1999(1+1999)+...+1999203(1+1999)
A=1999.2000+...+1999203.2000 chia hết cho 2000
A = 1999 + 19992 + ... + 19992044 (có 2044 số; 2044 chia hết cho 2)
A = (1999 + 19992) + (19993 + 19994) + ... + (19992043 + 19992044)
A = 1999.(1 + 1999) + 19993.(1 + 1999) + ... + 19992043.(1 + 1999)
A = 1999.2000 + 19993.2000 + ... + 19992043.2000
A = 2000.(1999 + 19993 + ... + 19992043) chia hết cho 2000
a: \(33^{44}>44^{33}>44^{32}\)
\(a,33^{44}=11^{44}\cdot3^{44}=11^{44}\cdot81^{11}>11^{33}\cdot64^{11}=11^{33}\cdot4^{33}=44^{33}>44^{32}\)
\(b,A=2000^{2016}\left(2000-1\right)+1999=1999\cdot2000^{2016}+1999⋮1999\)