So sánh:
a) A=20 + 21 + 22 + 23 + ... + 22010 và B= 22011 - 1.
b) A = 2009.2011 và B = 20102.
K
Khách
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18 tháng 11 2021
\(a,\Rightarrow2A=2+2^2+...+2^{2011}\)
\(\Rightarrow2A-A=2+2^2+...+2^{2011}-2^0-2-..-2^{2010}\)
\(\Rightarrow A=2^{2011}-1=B\)
\(b,A=2019.2011=\left(2010-1\right)\left(2010+1\right)=\left(2010-1\right).2010+\left(2010-1\right)=2010^2-2010+2010-1=2010^2-1< 2010^2=B\)
18 tháng 11 2021
\(a,\Rightarrow2A=2^1+2^2+...+2^{2011}\\ \Rightarrow2A-A=A=2^{2011}-2^0=2^{2011}-1=B\)
\(b,A=\left(2010-1\right)\left(2010+1\right)=2010^2+2010-2010-1=2010^2-1< 2010^2=B\)
HL
2
KV
14 tháng 11 2023
A = 2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰
⇒ 2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹
⇒ A = 2A - A = (2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹) - (2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰)
= 2²⁰¹¹ - 2⁰
= 2²⁰¹¹ - 1
= B
Vậy A = B
KV
26 tháng 12 2022
a) A = 2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰²²
2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²³
A = 2A - A
= (2 + 2² + 2³ + 2⁴ + ... + 2²⁰²³) - (2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰²²)
= 2²⁰²³ - 2⁰
= 2²⁰²³ - 1
Vậy A = B
b) A = 2021 . 2023
= (2022 - 1).(2022 + 1)
= 2022.(2022 + 1) - 2022 - 1
= 2022² + 2022 - 2022 - 1
= 2022² - 1 < 2022²
Vậy A < B
B
2
12 tháng 12 2021
Bài 1:
\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)
\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)
12 tháng 12 2021
Bài 2:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
2 tháng 12 2019
a/ \(2A=2+2^2+2^3+2^4+...+2^{2011}\)
\(A=2A-A=2^{2011}-2^0=2^{2011}-1=B\)
b/ \(A=2009.2011=\left(2010-1\right)\left(2010+1\right)=2010^2-1< B=2010^2\)
c/
\(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
\(\Rightarrow11^{24}=121^{12}< 125^{12}=5^{36}\)
d/
\(625^5=\left(5^4\right)^5=5^{20}\)
\(125^7=\left(5^3\right)^7=5^{21}>5^{20}=625^5\)
e/
\(3^{2n}=\left(3^2\right)^n=9^n\)
\(2^{3n}=\left(2^3\right)^n=8^n< 9^n=3^{2n}\)
f/
\(6.5^{22}>5.5^{22}=5^{23}\)
g/
\(333^{444}=\left(3.111\right)^{444}=3^{444}.111^{444}=\left(3^4\right)^{111}.111^{444}=81^{111}.111^{444}\)
\(444^{333}=\left(4.111\right)^{333}=4^{333}.111^{333}=\left(4^3\right)^{111}.111^{333}=64^{111}.111^{333}\)
\(\Rightarrow333^{444}>444^{333}\)
E
19 tháng 3 2021
Ta có :
\(A=2+2^2+2^3+2^4...2^{2010}\)\(^0\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(=2.3+2^3.3+....+2^{2009}.3\)
\(=3\left(2+2^3+....+2^{2009}\right)⋮3\)
Ta có :
\(2+2^2+2^3+2^4+....+2^{2010}\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(=2.7+2^4.7+....+2^{2008}.7\)
\(=7\left(2+2^4+....+2^{2008}\right)⋮7\)
Vậy \(2^1+2^2+2^3+2^4+...+2^{2010}⋮3\) và \(7\)
CT
0
a) \(A=2^0+2^1+2^2+...+2^{2010}\)
\(\Rightarrow2A=2^1+2^2+2^3+...+2^{2011}\)
\(\Rightarrow2A-A=\left(2^1+2^2+2^3+...+2^{2011}\right)-\left(2^0+2^1+2^2+...+2^{2010}\right)\)
\(\Rightarrow A=2^{2011}-2^0\)
\(\Rightarrow A=2^{2011}-1\)
Vì \(2^{2011}-1=2^{2011}-1\) nên \(A=B\)
Vậy A = B
b) Ta có: \(A=2009.2011=2009.\left(2010+1\right)=2009.2010+2009\)
\(B=2010^2=\left(2009+1\right).2010=2009.2010+2010\)
Vì \(2009.2010+2009< 2009.2010+2010\) nên A < B
Vậy A < B
\(A=2^0+2^1+2^2+2^3+....+2^{2010}\)
\(2.A=2\left(2^0+2^1+2^2+2^3+....+2^{2010}\right)\)
\(2.A=2.2^0+2.2+2.2^2+2.2^3+....+2.2^{2010}\)
\(2.A=2+2^2+2^3+2^4+....+2^{2011}\)
\(2A-A=\left(2+2^2+2^3+2^4+....+2^{2011}\right)-\left(2^0+2^1+2^2+2^3+....+2^{2010}\right)\)
\(A=\left(2-2^1\right)+\left(2^2-2^2\right)+\left(2^3-2^3\right)+....+\left(2^{2010}-2^{2010}\right)+2^{2011}-2^0\)
\(A=0+0+0+....+0+2^{2011}-2^0\)
\(A=2^{2011}-2^0\)
\(A=2^{2011}-1\)
Vì \(A=2^{2011}-1\) ; \(B=2^{2011}-1\)
\(=>A=B\)
Vậy \(A=B\)
b) \(A=2009.2001\)
\(A=\left(2010-1\right)\left(2010+1\right)\)
\(A=\left(2010-1\right).2010+\left(2010-1\right).1\)
\(A=2010.2010-2010.1+1.2010-1.1\)
\(A=2010^2-2010+2010-1\)
\(A=2010^2+0-1\)
\(A=2010^2-1\)
Vì \(A=2010^2-1\) ; \(B=2010^2\)
\(=>A< B\)
Vậy \(A< B\)