\(\dfrac{1}{2^{500}}=\dfrac{1}{\left(2^5\right)^{100}}=\dfrac{1}{32^{100}}\\ \dfrac{1}{5^{200}}=\dfrac{1}{\left(5^2\right)^{100}}=\dfrac{1}{25^{100}}\)
mà `32^(100)>25^(100)`
nên \(\dfrac{1}{2^{500}}>\dfrac{1}{5^{200}}\)
\(\dfrac{1}{2^{500}}=\dfrac{1}{\left(2^5\right)^{100}}=\dfrac{1}{32^{100}}\\ \dfrac{1}{5^{200}}=\dfrac{1}{\left(5^2\right)^{100}}=\dfrac{1}{25^{100}}\)
mà `32^(100)>25^(100)`
nên \(\dfrac{1}{2^{500}}>\dfrac{1}{5^{200}}\)
So sánh A và B :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
\(B=\dfrac{1}{2}\)
So sánh giá trị 2 biểu thức:
A=\(\dfrac{1+7+7^2+...+7^9}{1+7+7^2+...+7^{10}}\) và B=\(\dfrac{1+5+5^2+...+5^9}{1+5+5^2+...+5^{10}}\)
Mí bạn giúp tớ điii
giải rõ hộ nha :3
So sánh \(\dfrac{2019.2020-1}{2019.2020}\) và \(\dfrac{2020.2021-1}{2020.2021}\)
\(S=\dfrac{2}{2021+1}+\dfrac{2^2}{2021^2+1}+\dfrac{2^3}{2021^{2^2}+1}+...+\dfrac{2^{n+1}}{2021^{2^n}+1}+...+\dfrac{2^{2021}}{2021^{2^{2020}}+1}\)
So sánh S với \(\dfrac{1}{1010}\)
S=\(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}....\dfrac{199}{200}\)
CMR S2<\(\dfrac{1}{200}\)
So sánh : \(A=\dfrac{2019^{2020}+1}{2019^{2019}-1}\) và \(B=\dfrac{2019^{2019}+1}{2019^{2018}-1}\)
Cho tổng \(T=\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2020}{2^{2019}}+\dfrac{2021}{2^{2020}}\)
So sánh T với 3
So sánh A= \(\dfrac{10^{2023}+5}{10^{2022}+5}\) và B=\(\dfrac{10^{2022}+5}{10^{2021}+5}\)
So sánh A= \(\dfrac{10^{2023}+5}{10^{2022}+5}\) và B= \(\dfrac{10^{2022}+5}{10^{2022}+5}\)