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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
Ta có: \(B=2020.2021.2022=\left(2021-1\right).\left(2021+1\right).2021=\left(2021-1\right)^2.2021< 2021^2.2021=A\)
\(2.A=\frac{2^{2021}-2}{2^{2021}-1}=1-\frac{1}{2^{2021}-1}\)
\(2B=\frac{2^{2022}-2}{2^{2022}-1}=1-\frac{1}{2^{2022}-1}\)
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
Ta có : A = \(\frac{10^{2020}+1}{10^{2021}+1}\)
=> 10A = \(\frac{10^{2021}+10}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)
Lại có : \(B=\frac{10^{2021}+1}{10^{2022}+1}\)
=> \(10B=\frac{10^{2022}+10}{10^{2022}+1}=1+\frac{9}{10^{2022}+1}\)
Vì \(\frac{9}{10^{2022}+1}< \frac{9}{10^{2021}+1}\)
=> \(1+\frac{9}{10^{2022}+1}< 1+\frac{9}{10^{2022}+1}\)
=> 10B < 10A
=> B < A
b) Ta có : \(\frac{2019}{2020+2021}< \frac{2019}{2020}\)
Lại có : \(\frac{2020}{2020+2021}< \frac{2020}{2021}\)
=> \(\frac{2019}{2020+2021}+\frac{2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> \(\frac{2019+2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> B < A
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 $A> B$
Ta có: \(B=2020\cdot2021\cdot2022\)
\(=2021\left(2021-1\right)\left(2021+1\right)\)
\(=2021\left(2021^2-1\right)=2021^3-2021\)
=A-2021
=>A>B
ok
Đặt \(x = 2021^{2022}\) (rất lớn).
Khi đó:
\(a = \frac{x - 2018}{x - 2020} , b = \frac{x - 2020}{x - 2022}\)So sánh \(a\) và \(b\):
Xét:
\(a \textrm{ }\textrm{ } ? \textrm{ }\textrm{ } b \textrm{ }\textrm{ } \Longleftrightarrow \textrm{ }\textrm{ } \frac{x - 2018}{x - 2020} \textrm{ }\textrm{ } ? \textrm{ }\textrm{ } \frac{x - 2020}{x - 2022}\)Nhân chéo (vì các mẫu đều dương):
\(\left(\right. x - 2018 \left.\right) \left(\right. x - 2022 \left.\right) \textrm{ }\textrm{ } ? \textrm{ }\textrm{ } \left(\right. x - 2020 \left.\right)^{2}\)Khai triển:
- Vế trái:
\(\left(\right. x - 2018 \left.\right) \left(\right. x - 2022 \left.\right) = x^{2} - \left(\right. 2018 + 2022 \left.\right) x + 2018 \cdot 2022 = x^{2} - 4040 x + 4080396\)- Vế phải:
\(\left(\right. x - 2020 \left.\right)^{2} = x^{2} - 4040 x + 2020^{2} = x^{2} - 4040 x + 4080400\)So sánh:
\(4080396 < 4080400\)⇒
\(\left(\right. x - 2018 \left.\right) \left(\right. x - 2022 \left.\right) < \left(\right. x - 2020 \left.\right)^{2}\)⇒
\(a < b\)dc chua
ban
Ta có:
\(a=\frac{2021^{2022}-2018}{2021^{2022}-2020}=\frac{2021^{2022}-2020+2}{2021^{2022}-2020}=\frac{2021^{2022}-2020}{2021^{2022}-2020}+\frac{2}{2021^{2022}-2020}=1+\frac{2}{2021^{2022}-2020}\)
\(b=\frac{2021^{2022}-2020}{2021^{2022}-2022}=\frac{2021^{2022}-2022+2}{2021^{2022}-2022}=\frac{2021^{2022}-2022}{2021^{2022}-2022}+\frac{2}{2021^{2022}-2022}=1+\frac{2}{2021^{2022}-2022}\)
Ta thấy: \(2020<2021\rarr2021^{2022}-2020>2021^{2022}-2022\to\frac{2}{2021^{2022}-2020}<\frac{2}{2021^{2022}-2022}\to1+\frac{2}{2021^{2022}-2020}<1+\frac{2}{2021^{2022}-2022}\)
\(\to a<b\)
Vậy \(a<b\)