Cho A = \(1+2+2^2+...+2^{2021}\) và B = \(2^{2022}\). So sánh A và B.
Những câu hỏi liên quan
So sánh:
A = \(\dfrac{2^{2020}-1}{2^{2021}-1}\) và B = \(\dfrac{2^{2021}-1}{2^{2022}-1}\)
\(2.A=\frac{2^{2021}-2}{2^{2021}-1}=1-\frac{1}{2^{2021}-1}\)
\(2B=\frac{2^{2022}-2}{2^{2022}-1}=1-\frac{1}{2^{2022}-1}\)
dó \(\frac{1}{2^{2022}-1}< \frac{1}{2^{2021}-1}\Rightarrow1-\frac{1}{2^{2022}-1}>1-\frac{1}{2^{2021}-1}\Rightarrow A< B\)
HT
So sánh A và B : 1+2+2 mũ 2 +...+2 mũ 2021 + 2 mũ 2022 và B= 2 mũ 2023 -1 .
Ta có:
\(A=1+2+2^2+2^3+...+2^{2021}+2^{2022}\)
\(\Rightarrow2A=2\left(1+2+2^2+...+2^{2022}\right)\)
\(\Rightarrow2A=2+2^3+2^4+...+2^{2023}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2023}\right)-\left(1+2+2^2+...+2^{2022}\right)\)
\(\Rightarrow A=2^{2023}-1\)
Ta thấy: \(2^{2023}-1=2^{2023}-1\)
Vậy: \(A=B\)
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1. So sánh
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\) và B= \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{13}{60}\)
b) \(C=\dfrac{2019}{2021}+\dfrac{2021}{2022}\) và \(D=\dfrac{2020+2022}{2019+2021}.\dfrac{3}{2}\)
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
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so sánh a và b bt a= 2+2^2+2^3+....+2^2021
b=2^2022
A=2+22+23+...+22021
2A=22+23+24+...+22022
2A-A=(22+23+24+...+22022)-(2+22+23+...+22021)
A=22022-2 mà B= 22022 nên A<B.
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so sánh A và B
biết A=1+3+3^2+3^3+....+3^2021 B=(3^2022-1):2
A=1+3+32+33+.....+32021
-->3A=3(1+3+32+33+.....+32021)
-->3A=3+32+33+...+32022
-->3A-A=(3+32+33+....32022)-(1+3+32+33+.....+32021)
-->2A=32022-1
-->A=(32022-1):2
Vì (32022-1):2>(32022-1):2
-->A=B
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so sánh b=1/2022+2/2021+3/2020+...+2021/2+2022/1 VÀ c=1/2+1/3+1/4+...+1/2022+1/2023
B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)
B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022
B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\)
B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\)
B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))
Vậy B > C
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so sánh A và B:
a) A= 2 + 2^2 + 2^3 +...+ 2^2021 và B= 2^2022
b) A= 5^200 và B= 2^500
\(A=2+2^2+2^3+...+2^{2021}\)
\(2A=2^2+2^3+2^4+...+2^{2021}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2021}\right)-\left(2+2^2+2^3+...+2^{2021}\right)\)
\(A=2^2+2^3+2^4+...+2^{2021}-2-2^2-2^3-...-2^{2021}\)
\(A=2^{2021}-2\)
So sánh A=2021/2022 và B=1/1.2+1/2.3+1/3.4+2...+1/97.98+1/98.99
\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}\)
\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{99-98}{98.99}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
\(=1-\dfrac{1}{99}\)
\(A=\dfrac{2021}{2022}=\dfrac{2022-1}{2022}=1-\dfrac{1}{2022}\)
Có \(2022>99>0\Leftrightarrow\dfrac{1}{99}>\dfrac{1}{2022}\)
Suy ra \(A>B\).
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So sánh
A = 2 + 2 mũ 2 + 2 mũ 3 + 2 mũ 4 +....+ 2 mũ 2021 và B = 2 mũ 2022
\(A=2+2^2+2^3+...+2^{2021}\\ \Leftrightarrow2A=2^2+2^3+2^4+...+2^{2022}\\ \Leftrightarrow2A-A=\left(2^2+2^3+2^4+...+2^{2022}\right)-\left(2+2^2+2^3+...+2^{2021}\right)\\ \Leftrightarrow A=2^{2022}-2\\ 2^{2022}-2< 2^{2022}\Rightarrow A< B\)
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A = 2 + 2 2 + 2 3 + . . . + 2 2021 ⇔ 2 A = 2 2 + 2 3 + 2 4 + . . . + 2 2022 ⇔ 2 A − A = ( 2 2 + 2 3 + 2 4 + . . . + 2 2022 ) − ( 2 + 2 2 + 2 3 + . . . + 2 2021 ) ⇔ A = 2 2022 − 2 2 2022 − 2 < 2 2022 ⇒ A < B
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