BQ
1
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Khách
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9 tháng 10 2018
\(C=\left(2018^{2019}+2018^{2018}+...+2018^2+2018\right)2017+1\)
\(=\left(2018^{2019}+2018^{2018}+...+2018^2+2018\right)2018-\left(2018^{2019}+2018^{2018}+...+2018\right)-1\)
\(=\left(2018^{2020}+2018^{2019}+...+2018^3+2018^2\right)-\left(2018^{2019}+2018^{2018}+...+2018^2+2018\right)+1\)\(=2018^{2020}-2018+1\)
\(=2018^{2020}-2017\)
BM
1
15 tháng 10 2018
\(M=\left(2018^{2019}+2018^{2018}+...+2018^2+2018\right)2017+1\)
Gọi \(A=2018^{2019}+2018^{2018}+...+2018^2+2018\)
\(\Rightarrow2018A=2018^{2020}+2018^{2019}+...+2018^3+2018^2\)
\(\Rightarrow2018A-A=2018^{2020}-2018\)
\(\Rightarrow2017A=2018^{2020}-2018\)
\(\Rightarrow A=\left(2018^{2020}-2018\right)\div2017\)
\(\Rightarrow M=\left(2018^{2020}-2018\right)\div2017.2017+1\)
\(\Rightarrow M=2018^{2020}-2018+1\)
\(\Rightarrow M=2018^{2020}-2017\)
NM
2
14 tháng 10 2018
A=22019-(22018+22017+...+21+20)
Đặt M =22018+22017+...+21+20
M=22018+22017+...+2+1
2M=22019+22018+...+22+2
2M-M=(22019+22018+...+22+2)-(22018+22017+...+2+1)
M=22019-1
Suy ra:A=22019-(22019-1)
A=22019-22019+1
A=1
Vậy A=1
14 tháng 10 2018
Ta có : \(A=2^{2019}-\left(2^{2018}+2^{2017}+...+2^1+2^0\right)\)
Đặt \(B=2^0+2^1+...+2^{2017}+2^{2018}\\ \Rightarrow2B=2+2^2+...+2^{2019}\\ \Rightarrow2B-B=\left(2+2^2+...+2^{2019}\right)-\left(2^0+2^1+...+2^{2017}+2^{2018}\right)\\ \Rightarrow B=2^{2019}-2^0\\ \Rightarrow A=2^{2019}-\left(2^{2019}-2^0\right)\\ \Rightarrow A=2^0=1\)
Vậy A=1
NM
Nguyễn Minh Quang
Giáo viên
24 tháng 7 2021
a. ta có \(3^{102}=3^{3\times34}=27^{34}>25^{34}=5^{2\times34}=5^6\text{ vậy }3^{102}>5^{68}\)
b. ta có \(C=1+2+..+2^{2017}\text{ nên }2C=2+2^2+...+2^{2018}\)
lấy hiệu ta có : \(C=\left(2+2^2+..+2^{2018}\right)-\left(1+2+..+2^{2017}\right)=2^{2018}-1< 2^{2018}\)
Vậy \(C< 2^{2018}\)
c. dễ thấy \(C>\frac{1}{2}=F\)
d. ta có \(5G=1+\frac{1}{5}+..+\frac{1}{5^{2016}}\Rightarrow4G=1-\frac{1}{5^{2017}}\)hay \(G=\frac{1}{4}-\frac{1}{4\times5^{2017}}< \frac{1}{4}=H\text{ hay }G< H\)
16 tháng 8 2017
Ta có \(A=1+2+2^2+2^3+...+2^{2017}\)
Suy ra\(2.A=2+2^2+2^3+2^4+....+2^{2018}\)
Khi đó \(2A-A=2+2^2+2^3+2^4+....+2^{2018}-\left(1+2+2^2+2^3+....+2^{2017}\right)\)
Hay \(A=2^{2018}-1\)
Ta thấy \(A=2^{2018}-1\); \(B=2^{2018}-1\)nên \(A=B\)
Vậy \(A=B\)
2 tháng 1 2020
\(x=\frac{2019^{2020}+1}{2019^{2019}+1}>\frac{2019^{2020}+1+2018}{2019^{2019}+1+2018}=\frac{2019^{2020}+2019}{2019^{2019}+2019}=\frac{2019\left(2019^{2019}+1\right)}{2019\left(2019^{2018}+1\right)}=\frac{2019^{2019}+1}{2019^{2018}+1}\)(1)
\(y=\frac{2019^{2019}+2020}{2019^{2018}+2020}< \frac{2019^{2019}+2020-2019}{2019^{2018}+2020-2019}=\frac{2019^{2019}+1}{2019^{2018}+1}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow x>y\)
Đặt \(A=\frac{2^{2017}+1}{2^{2018}+1}\Rightarrow2A=\frac{2^{2018}+2}{2^{2018}+1}=\frac{2^{2018}+1+1}{2^{2018}+1}=1+\frac{1}{2^{2018}+1}\)
\(B=\frac{2^{2018}+1}{2^{2019}+1}\Rightarrow2B=\frac{2^{2019}+2}{2^{2019}+1}=\frac{2^{2019}+1+1}{2^{2019}+1}=1+\frac{1}{2^{2019}+1}\)
Vì \(2^{2019}+1>2^{2018}+1\Rightarrow\frac{1}{2^{2019}+1}< \frac{1}{2^{2018}+1}\)
\(\Rightarrow2A>2B\Rightarrow A>B\)