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a) \(\dfrac{3}{4}=\dfrac{3x4}{4x4}=\dfrac{12}{16},\dfrac{6}{7}=\dfrac{6x2}{7x2}=\dfrac{12}{14}\)
Do 16 > 14 => \(\dfrac{12}{16}< \dfrac{12}{14}hay\dfrac{3}{4}< \dfrac{6}{7}\)
a/
2020.2021=(2019+1)(2022-1)=
=2019.2022-2019+2022-1=2019.2022+2>2019.2022
b/
\(4^7=\left(2^2\right)^7=2^{14}< 2^{15}\)
c/
\(199^{20}< 200^{20}=\left(8.25\right)^{20}=\left(2^3.5^2\right)^{20}=2^{60}.5^{40}\)
\(2000^{15}=\left(16.125\right)^{15}=\left(2^4.5^3\right)^{15}=2^{60}.5^{45}\)
\(\Rightarrow2000^{15}=2^{60}.5^{45}>2^{60}.5^{40}>199^{20}\)
d/
\(31^{31}< 32^{31}=\left(2^5\right)^{31}=2^{155}\)
\(17^{39}>16^{39}=\left(2^4\right)^{39}=2^{156}\)
\(\Rightarrow17^{39}=2^{156}>2^{155}>31^{31}\)
\(\dfrac{-11}{-32}>\dfrac{16}{49}\)
\(\dfrac{-2020}{-2021}>\dfrac{-2021}{2022}\)
a, \(\frac{15}{106}\)và \(\frac{21}{133}\)
Ta có:
\(\frac{15}{106}< \frac{15}{100}=\frac{3}{20}=\frac{21}{140}< \frac{21}{133}\)
\(\Rightarrow\frac{15}{106}< \frac{21}{133}\)
Vậy ........
b, \(\frac{31}{100}\)và \(\frac{89}{150}\)
Ta có:
\(\frac{31}{100}< \frac{31}{93}=\frac{1}{3}=\frac{50}{150}< \frac{89}{150}\)
\(\Rightarrow\frac{31}{100}< \frac{89}{150}\)
Vậy........
c, \(\frac{2020}{2019}\)và \(\frac{2021}{2020}\)
Ta có:
\(\frac{2020}{2019}-1=\frac{1}{2019}\) ;
\(\frac{2021}{2020}-1=\frac{1}{2020}\)
Vì \(\frac{1}{2019}>\frac{1}{2020}\)
\(\Rightarrow\frac{2020}{2019}-1>\frac{2021}{2020}-1\)
\(\Rightarrow\frac{2020}{2019}>\frac{2021}{2020}\)
Vậy .........
d, n+2019/n+2021 và n+2020/n+2022
Câu d bn tự lm nhé
a) <
b) >
c) <
d) >
Thanks bro