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a) Ta có: 2003^152>2003^20>199^20
Vậy 2003^152>199^20
b) Ta có: 3^39=(3^13)^3=1594323^3
11^21=(11^7)^3=19487171^3
Vì 1594323^3<19487171^3 nên 3^39<11^21
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+2+2^2+2^3+...+2^9\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{10}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+2^4+...+2^{10}\right)-\left(1+2+2^2+2^3+...+2^9\right)\)
\(\Rightarrow A=2^{10}-1\)
Ta có :
\(5.2^8=4.2^8+1.2^8=2^2.2^8+2^8=2^{10}+2^8\)
\(\Rightarrow A< 5.2^8\)
Vậy A < 5 . 28
Câu hỏi của Nguyễn Anh Thư - Toán lớp 6 - Học toán với OnlineMath
Bạn tham khảo nhé
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Ta có: A= 1+2+2^2+2^3+...+2^2018
2A = 2+2^2+2^3+2^4+...+2^2019
2A-A=A= 2^2019-1 = (2^2017.4) -1
Mà B=5.2^2017
=> (2^2017.4) -1 < 5.2^2017
=> A < B
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\(2A=2+2^2+...+2^{10}\)
\(2A-A=\left(2+2^2+...+2^{10}\right)-\left(1+2+...+2^9\right)\)
\(A=2^{10}-1=1023\)
mà \(5\cdot2^8=1280\Rightarrow A< 5\cdot2^8\)
A = 1 + 2+22+23+...+29
2A = 2 + \(2^2+2^3+2^4+...+2^{10}\)
\(-\)
\(A=1+2+2^2+2^3+...+2^9\)
\(A=\)\(1-2^{10}\)
KL= tự so sánh nha
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a) S= 1+2+22+...+29
2S=2+22+23+...+210
2S-S=(2+22+23+...+210)-(1+2+23+...+29)
S=210-1
5.28=2.2+1.28=1+22.28=1+210
=>S=5.28
b) A=1+2+22+....+2100
2A=2+22+23+...+2101
2A-A=(2+22+23+...+2101)-(1+2+22+...+2100)
A=2101-1
=> A<2101
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