So sánh: \(\log_{2019}2020\) và \(\log_{2020}2021\)
Những câu hỏi liên quan
rút gọn các biểu thức
a) \(log_{a^4}b^4.log_ba^5\)
b) \(log_{a^3}b^2.log_ba^4\)
c) \(log_{a^{15}}b^7.log_{b^{49}}a^{30}\)
d) \(log_{a^{2021}}b^{2020}.log_{b^{4040}}a^{6063}\)
\(log_{a^4}b^4.log_ba^5=\dfrac{1}{4}.4.log_ab.5.log_ba=5.log_ab.log_ba=5\)
\(log_{a^3}b^2.log_ba^4=\dfrac{1}{3}.2.log_ab.4.log_ba=\dfrac{8}{3}.log_ab.log_ba=\dfrac{8}{3}\)
\(log_{a^{15}}b^7.log_{b^{49}}a^{30}=\dfrac{1}{15}.7.log_ab.\dfrac{1}{49}.30.log_ba=\dfrac{2}{7}log_ab.log_ba=\dfrac{2}{7}\)
\(log_{a^{2021}}b^{2020}.log_{b^{4040}}a^{6063}=\dfrac{1}{2021}.2020.log_ab.\dfrac{1}{4040}.6063.log_ba=\dfrac{3}{2}\)
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so sánh P=2019/2020+2020/2021+2021/2022 và Q=2019+2020+2021/2020+2021+2022
17 tháng 4 2021 lúc 17:53
So sánh M = \(\dfrac{2019}{2020}+\dfrac{2020}{2021}\) và N = \(\dfrac{2019+2020}{2020+2021}\)
Giải:
Ta có: N=2019+2020/2020+2021
=>N=2019/2020+2021 + 2020/2020+2021
Vì 2019/2020 > 2019/2020+2021 ; 2020/2021 > 2020/2020+2021
=>M>N
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Ta có : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)
\(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)
\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)
\(\Rightarrow M>N\)
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2019 x 2020 - 1/ 2019 x 2020 và 2020 x 2021 - 1/ 2020 x 2021
so sánh phân số
So sánh:
\(A=\frac{2019}{2020}+\frac{2020}{2021}\) và \(B=\frac{2019+2020}{2020+2021}\)
Ta có: \(\frac{2019}{2020}>\frac{2019}{2020+2021};\frac{2020}{2021}>\frac{2020}{2020+2021}\)
=> \(\frac{2019}{2020}+\frac{2020}{2021}>\frac{2019}{2020+2021}+\frac{2020}{2020+2021}=\frac{2019+2020}{2020+2021}\)
=> A > B.
so sánh: A=2019^2019+1/2019^2020+1 và B=2019^2020+1/2019^2021+1
Vì 2019 + 2020 < 2019 + 2021 nên A < B
so sánh: -2019/2019 và -2021/2020
\(\dfrac{-2019}{2019}=-1\)
\(\dfrac{-2021}{2020}=-1,004\)
\(\Rightarrow\dfrac{-2019}{2019}>\dfrac{-2021}{2020}\)
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\(-\dfrac{2019}{2019}=-1\)
\(\dfrac{-2021}{2020}< -1\)
Do đó: \(-\dfrac{2019}{2019}>\dfrac{-2021}{2020}\)
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12 tháng 2 2023 lúc 17:57
So sánh A=\(\dfrac{2018}{2019}\)+\(\dfrac{2019}{2020}\)+\(\dfrac{2020}{2021}\)+\(\dfrac{2021}{2018}\)với 4
Lời giải:
$A=1-\frac{1}{2019}+1-\frac{1}{2020}+1-\frac{1}{2021}+1+\frac{3}{2018}$
$=4+(\frac{1}{2018}-\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2020}+\frac{1}{2018}-\frac{1}{2021})$
$> 4+0+0+0+0=4$
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So sánh 2018/2019 + 2019/2020 + 2020/2021:3
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