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21 tháng 9 2021

 điền dấu <

21 tháng 9 2021

Ta có : \(\frac{2020}{2021}=\frac{2020.2022}{2021.2022}=\frac{\left(2021-1\right)\left(2021+1\right)}{2021.2022}=\frac{2021^2-1}{2021.2022}\)

\(\frac{2021}{2022}=\frac{2021^2}{2021.2022}\)

Vì 20212 > 20212 - 1 nên \(\frac{2021^2-1}{2021.2022}< \frac{2021^2}{2021.2022}\)

Hay \(\frac{2020}{2021}< \frac{2021}{2022}\)

Tham khảo:

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17 tháng 1 2022

\(\dfrac{2021}{2022}=\dfrac{2020}{2021}\)

17 tháng 1 2022

\(\dfrac{2021}{2022}\) và \(\dfrac{2020}{2021}\)

\(\dfrac{2021}{2022}=1-\dfrac{1}{2022}\)

\(\dfrac{2020}{2021}=1-\dfrac{1}{2021}\)

\(\text{Vì }\)\(\dfrac{1}{2022}>\dfrac{1}{2021}=>1-\dfrac{1}{2022}>1-\dfrac{1}{2021}=>\dfrac{2021}{2022}>\dfrac{2020}{2021}\)

26 tháng 9 2021

Ta có: \(B=2020.2021.2022=\left(2021-1\right).\left(2021+1\right).2021=\left(2021-1\right)^2.2021< 2021^2.2021=A\)

3 tháng 5 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 

 

1 tháng 12 2023

Có: \(2022>2020\)

\(\Rightarrow\dfrac{1}{2022}< \dfrac{1}{2020}\)

\(\Rightarrow\dfrac{2021}{2022}< \dfrac{2021}{2020}\)

\(2.A=\frac{2^{2021}-2}{2^{2021}-1}=1-\frac{1}{2^{2021}-1}\)

\(2B=\frac{2^{2022}-2}{2^{2022}-1}=1-\frac{1}{2^{2022}-1}\)

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

28 tháng 5 2022

1) \(16^{2020}+\dfrac{1}{16^{2021}}+1\)

\(=16^{2021}\div16^{2020}+1\)

\(=16+1\)

\(=17\)

2) \(16^{2021}+\dfrac{1}{16^{2022}}+1\)

\(=16^{2022}\div16^{2021}+1\)

\(=16+1\)

= 17

Vì 17=17 nên \(16^{2020}+\dfrac{1}{16^{2021}}+1=16^{2021}+\dfrac{1}{16^{2022}}+1\)

27 tháng 8 2021

\(\left(2022+2021\right)^{2020}>\left(1998+1997\right)^{2020}>\left(1998+1997\right)^{1996}\)

tick mik nha

27 tháng 8 2021

(2022+2021)>(1998+1997)

AH
Akai Haruma
Giáo viên
30 tháng 4 2023

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 $A> B$