Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: 43/52>26/52=1/2=60/120
b: 17/68=1/4<1/3=35/105<35/103
c: \(\dfrac{2018\cdot2019-1}{2018\cdot2019}=1-\dfrac{1}{2018\cdot2019}\)
\(\dfrac{2019\cdot2020-1}{2019\cdot2020}=1-\dfrac{1}{2019\cdot2020}\)
2018*2019<2019*2020
=>-1/2018*2019<-1/2019*2020
=>\(\dfrac{2018\cdot2019-1}{2018\cdot2019}< \dfrac{2019\cdot2020-1}{2019\cdot2020}\)
Ta thấy : \(2222^{3333}vs2^{300}:\hept{\begin{cases}2222>2\\3333>300\end{cases}\Rightarrow2222^{3333}>2^{300}}\)
Ta thấy : \(2222^{1111}=1111^{1111}.2^{1111}< 1111^{1111}.1111^{1110}=1111^{2221}\)
Ta thấy : \(54^{10}=\left(3^3\right)^{10}.2^{10}=3^{30}.2^{10}=3^{12}.3^{18}.2^{10}>3^{12}.7^{12}=21^{12}.\)
Ta có : \(N=2022.2024\)
\(N=\left(2023-1\right)\left(2023+1\right)\)
\(N=2023^2+2023-2023-1\)
\(N=2023^2-1\)
Mà : \(M=2023.2023=2023^2\)
\(\Rightarrow M>N\)
a) ta có: \(1-\frac{2012}{2013}=\frac{1}{2013}\)
\(1-\frac{2013}{2014}=\frac{1}{2014}\)
mà \(\frac{1}{2013}>\frac{1}{2014}\) nên \(\frac{2013}{2014}>\frac{2012}{2013}\)
\(\dfrac{19}{19}\) = 1 < \(\dfrac{2005}{2004}\) vậy \(\dfrac{19}{19}\) < \(\dfrac{2005}{2004}\)
\(\dfrac{72}{73}\) = 1 - \(\dfrac{1}{73}\)
\(\dfrac{98}{99}\) = 1 - \(\dfrac{1}{99}\)
Vì \(\dfrac{1}{73}\) > \(\dfrac{1}{99}\) nên \(\dfrac{72}{73}\) < \(\dfrac{98}{99}\)
\(201^{60}=\left(201^4\right)^{15}=1944810000^{15}\)
\(398^{45}=\left(398^3\right)^{15}=63044792^{15}\)
Do \(1944810000>63044792\)
\(\Rightarrow1944810000^{15}>63044792^{15}\)
\(\Rightarrow201^{60}>398^{45}\)
Ta có:
\(201^{60}>200^{60};398^{45}< 400^{45}\)
\(200^{60}=\left(2.100\right)^{60}=2^{60}.100^{60}=2^{60}.\left(10^2\right)^{60}\)
\(=2^{60}.10^{120}=2^{60}.10^{30}.10^{90}\)
\(400^{45}=\left(2.100\right)^{45}=2^{45}.100^{45}=2^{45}.\left(10^2\right)^{45}\)
\(=2^{45}.10^{90}\)
Mà \(2^{60}.10^{30}.10^{90}>2^{45}.10^{90}\)
\(\Rightarrow200^{60}>400^{45}\)
\(\Rightarrow201^{60}>200^{60}>400^{45}>398^{45}\)
\(\Rightarrow201^{60}>398^{45}\)
> (mik hk viết lũy thừa chắc bn hỉu ùi) k nha