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   \(\frac{2013}{2014}>\frac{2012}{2013}\)

sao giống lớp 4 thế ta

\(\dfrac{19}{19}\) = 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 \(\dfrac{72}{73}\) < \(\dfrac{98}{99}\) 

1) 19/19 < 2005/2004

2)72/73 > 98/99

giúp mình giải câu này với các bạn khó quá

Ta có : 10 = 10

-> 10*A = 10*B

Từ đó -> A = B

\(201^{60}=\left(201^4\right)^{15}=1944810000^{15}\)

\(398^{45}=\left(398^3\right)^{15}=63044792^{15}\)

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 \(2^{60}.10^{30}.10^{90}>2^{45}.10^{90}\)

\(\Rightarrow200^{60}>400^{45}\)

\(\Rightarrow201^{60}>200^{60}>400^{45}>398^{45}\)

\(\Rightarrow201^{60}>398^{45}\)

8>32 hok tút

8 va 32

32 lon hon 8

8 be hon 32

Ta có: 37/-49 < 0; -12/-35 > 0

Suy ra 37/-49 < -12/-35

Ta có:

\(\dfrac{37}{-49}< 0;\dfrac{-12}{-35}=\dfrac{12}{35}>0\)

\(\Rightarrow\dfrac{37}{-49}< \dfrac{-12}{-35}\)

Vậy...