C = \(\dfrac{10^{2021}+1}{10^{2020}+1}\) và D = \(\dfrac{10^{2022}+1}{10^{2021}+1}\)
Những câu hỏi liên quan
Cho A = \(\dfrac{10^{2020}-1}{10^{2021}-1}\) và B = \(\dfrac{10^{2021}+1}{10^{2022}+1}\)
So sánh A và B
Lời giải:
$10A=\frac{10^{2021}-10}{10^{2021}-1}=\frac{10^{2021}-1-9}{10^{2021}-1}$
$=1-\frac{9}{10^{2021}-1}>1$
$10B=\frac{10^{2022}+10}{10^{2022}+1}=\frac{10^{2022}+1+9}{10^{2022}+1}$
$=1+\frac{9}{10^{2022}+1}<1$
$\Rightarrow 10A> 1> 10B$
Suy ra $A> B$
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So sánh A và B:
A= \(\dfrac{10^{2020}+1}{10^{2021}+1}\) B=\(\dfrac{10^{2021}+1}{10^{2022}+1}\)
Giúp mình với!
Ta có:
\(10A=\dfrac{10\left(10^{2020}+1\right)}{10^{2021}+1}=\dfrac{10^{2021}+10}{10^{2021}+1}=1+\dfrac{9}{10^{2021}+1}\)
\(10B=\dfrac{10\left(10^{2021}+1\right)}{10^{2022}+1}=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
⇒ \(10A>10B\) ( vì \(\dfrac{9}{10^{2021}+1}>\dfrac{9}{10^{2022}+1}\) )
Suy ra: \(A>B\)
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a) A = \(\dfrac{10^{2020}+1}{10^{2021}+1}\) và B = \(\dfrac{10^{2021}+1}{10^{2022}+1}\). So sánh A và B
b) Vẽ 10 đường thẳng cùng đi qua 1 điểm A. Tính số góc đỉnh A tạo thành.
Áp dụng tính chất : Nếu \(\dfrac{a}{b}\) < 1 thì \(\dfrac{a}{b}\) < \(\dfrac{a+n}{b+n}\) ( a ϵ N; b; n ϵ N* )
Ta có \(B=\dfrac{10^{2021}+1}{10^{2022}+1}< \dfrac{10^{2021}+10}{10^{2022}+10}=\dfrac{10\left(10^{2020}+1\right)}{10\left(10^{2021}+1\right)}=\dfrac{10^{2020}+1}{10^{2021}+1}=A\)
Vậy A > B
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A = \(\dfrac{10^{2020}+1}{10^{2021}+1}\) ⇒ 10\(\times\) A = \(\dfrac{10^{2020}+1}{10^{2021}+1}\) \(\times\) 10
10A = \(\dfrac{10^{2021}+10}{10^{2021}+1}\) =1+\(\dfrac{9}{10^{2021}+1}\)
B = \(\dfrac{10^{2021}+1}{10^{2022}+1}\) ⇒ 10 \(\times\) B = \(\dfrac{10^{2021}+1}{10^{2022}+1}\) \(\times\) 10
10B = \(\dfrac{10^{2022}+10}{10^{2022}+1}\) = 1 + \(\dfrac{9}{10^{2022}+1}\)
Vì \(\dfrac{9}{10^{2021}+1}\) > \(\dfrac{9}{10^{2022}+1}\)
Vậy 10A > 10B ⇒ A > B
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12 tháng 3 2023 lúc 22:57
So sánh:
A=\(\dfrac{10^{2022}+1}{10^{2023}+1}\) và B=\(\dfrac{10^{2021}+1}{10^{2022}+1}\)
\(10A=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)
\(10B=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
2023>2022
=>10^2023+1>10^2022+1
=>10A<10B
=>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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3 tháng 5 2022 lúc 12:31
a) tìm x :\(\dfrac{2}{1.4}x+\dfrac{2}{4.7}x+\dfrac{2}{7.10}x+....+\dfrac{2}{31.344}x=10\)
b)so sánh hai phân số sau : A=\(\dfrac{6^{2020}+1}{6^{2021}+1}\)và B=\(\dfrac{6^{\text{2021}}+1}{\text{6}^{\text{2022}}+1}\)
ét o ét giúp với ạ
so sánh:
a)C= \(\dfrac{100^{99}+1}{100^{100}+1}\) và D= \(\dfrac{100^{100}+1}{100^{101}+1}\)
b)E=\(\dfrac{2020^{2021}+1}{2020^{2022}+1}\) và F=\(\dfrac{2020^{2020}+1}{2020^{2021}+1}\)
c: \(100C=\dfrac{100^{100}+100}{100^{100}+1}=1+\dfrac{99}{100^{100}+1}\)
\(100D=\dfrac{100^{101}+100}{100^{101}+1}=1+\dfrac{99}{100^{101}+1}\)
100^100+1<100^101+1
=>\(\dfrac{99}{100^{100}+1}>\dfrac{99}{100^{101}+1}\)
=>100C>100D
=>C>D
b: \(2020E=\dfrac{2020^{2022}+2020}{2020^{2022}+1}=1+\dfrac{2019}{2020^{2022}+1}\)
\(2020F=\dfrac{2020^{2021}+2020}{2020^{2021}+1}=1+\dfrac{2019}{2020^{2021}+1}\)
2020^2022+1>2020^2021+1(Do 2022>2021)
=>\(\dfrac{2019}{2020^{2022}+1}< \dfrac{2019}{2020^{2021}+1}\)
=>2020E<2020F
=>E<F
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Bài 1: So sánh phấ số
A= 2020/2021 + 2021/2022 và B= 2020+2021/2021+2022
Bài 2: So Sánh phân số
A= 10^10 -1 / 10^11 -1 va B= 10^10 +1 / 10^11 +1
A = \(\dfrac{2020}{2021}\) + \(\dfrac{2021}{2022}\) ; B = \(\dfrac{2020+2021}{2021+2022}\)
B = \(\dfrac{2020+2021}{2021+2022}\) = \(\dfrac{2020}{2021+2022}\) + \(\dfrac{2021}{2021+2022}\)
\(\dfrac{2020}{2021}\) > \(\dfrac{2020}{2021+2022}\)
\(\dfrac{2021}{2022}\) > \(\dfrac{2021}{2021+2022}\)
Cộng vế với vế ta có:
A = \(\dfrac{2020}{2021}\) + \(\dfrac{2021}{2022}\) > \(\dfrac{2020}{2021+2022}\) + \(\dfrac{2021}{2021+2022}\) = B
Vậy A > B
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A = \(\dfrac{10^{10}-1}{10^{11}-1}\)
A \(\times\) 10 = \(\dfrac{(10^{10}-1)\times10}{10^{11}-1}\) = \(\dfrac{10^{11}-10}{10^{11}-1}\) = 1 - \(\dfrac{9}{10^{11}-1}\) < 1
B = \(\dfrac{10^{10}+1}{10^{11}+1}\)
B \(\times\) 10 = \(\dfrac{(10^{10}+1)\times10}{10^{11}+1}\) = \(\dfrac{10^{11}+10}{10^{11}+1}\) = 1 + \(\dfrac{9}{10^{11}+1}\) > 1
Vì 10 A< 1< 10B
Vậy A < B
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Thực hiện phép tính:
a) 2021 - \(\left(\dfrac{1}{3}\right)^2\) . 32
b \(\dfrac{5}{10}\) + 9 . \(\dfrac{-3}{2}\)
c) -10 . \(\left(-\dfrac{2021}{2022}\right)^0\) + \(\left(\dfrac{2}{5}\right)^2\) : 2
a) 2021 - (1/3)² . 3²
= 2021 - 1/9 . 9
= 2021 - 1
= 2020
b) 5/10 + 9 . (-3/2)
= 1/2 - 27/2
= -26/2
= -13
c) -10 . (-2021/2022)⁰ + (2/5)² : 2
= -10 . 1 + 4/25 . 2
= -10 + 8/25
= -68/7
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\(a,2021-\left(\dfrac{1}{3}\right)^2\cdot3^2\\ =2021-\dfrac{1}{9}\cdot9\\ =2021-\dfrac{9}{9}\\ =2021-1=2020\\ b,\dfrac{5}{10}+9\cdot\dfrac{-3}{2}\\ =\dfrac{5}{10}+\dfrac{-27}{2}\\ =\dfrac{5}{10}+\dfrac{-135}{10}\\ =-\dfrac{130}{10}\\ =-13\\ c,-10\cdot\left(-\dfrac{2021}{2022}\right)^0+\left(\dfrac{2}{5}\right)^2:2\\ =-10\cdot1+\dfrac{4}{25}\cdot\dfrac{1}{2}\\ =-10+\dfrac{4}{50}\\ =-10+\dfrac{2}{25}\\ =-\dfrac{248}{25}\)
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