\(C=\dfrac{2^{2024}-3}{2^{2023}-1}=\dfrac{2.2^{2023}-2-1}{2^{2023}-1}=\dfrac{2\left(2^{2023}-1\right)-1}{2^{2023}-1}=2-\dfrac{1}{2^{2023}-1}\)

\(D=\dfrac{2^{2023}-3}{2^{2022}-1}=\dfrac{2.2^{2022}-2-1}{2^{2022}-1}=\dfrac{2\left(2^{2022}-1\right)-1}{2^{2022}-1}=2-\dfrac{1}{2^{2022}-1}\)

Ta có

\(2^{2023}>2^{2022}\Rightarrow2^{2023}-1>2^{2022}-1\)

\(\Rightarrow\dfrac{1}{2^{2023}-1}< \dfrac{1}{2^{2022}-1}\Rightarrow2-\dfrac{1}{2^{2023}-1}>2-\dfrac{1}{2^{2022}-1}\)

\(\Rightarrow C>D\)

Lời giải:
Xét hiệu: 

$\frac{2022}{\sqrt{2023}}+\frac{2023}{\sqrt{2022}}-(\sqrt{2022}+\sqrt{2023})$

$=(\frac{2022}{\sqrt{2023}}-\sqrt{2023})+(\frac{2023}{\sqrt{2022}}-\sqrt{2022})$

$=\frac{2022-2023}{\sqrt{2023}}+\frac{2023-2022}{\sqrt{2022}}$

$=\frac{1}{\sqrt{2022}}-\frac{1}{\sqrt{2023}}>0$

$\Rightarrow \frac{2022}{\sqrt{2023}}+\frac{2023}{\sqrt{2022}}>\sqrt{2022}+\sqrt{2023}$

-2024/2023<-1

-1<-2023/2024

=>-2024/2023<-2023/2024

Áp dụng tính chất : Nếu \(\dfrac{a}{b}< 1\) thì \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\) ( a; b; n ϵ N , b; n ≠ 0 )

Ta có \(\dfrac{2023^{31}+5}{2023^{32}+5}< 1\)

⇒ \(B=\dfrac{2023^{31}+5}{2023^{32}+5}< \dfrac{2023^{31}+5+2018}{2023^{32}+5+2018}=\dfrac{2023^{31}+2023}{2023^{32}+2023}=\dfrac{2023\left(2023^{30}+1\right)}{2023\left(2023^{31}+1\right)}=\dfrac{2023^{30}+1}{2023^{31}+1}=A\)Vậy A > B

Ta có 2023A = \(\dfrac{2023.\left(2023^{30}+5\right)}{2023^{31}+5}=\dfrac{2023^{31}+5.2023}{2023^{31}+5}\)

\(=1+\dfrac{2022.5}{2023^{31}+5}\)

Lại có 2023B = \(\dfrac{2023.\left(2023^{31}+5\right)}{2023^{32}+5}=\dfrac{2023^{32}+2023.5}{2023^{32}+5}\)

\(=1+\dfrac{2022.5}{2023^{32}+5}\)

Dễ thấy 2023³¹ + 5 < 2023³² + 5

\(\Leftrightarrow\dfrac{2022.5}{2023^{31}+5}>\dfrac{2022.5}{2023^{32}+5}\)

\(\Leftrightarrow1+\dfrac{2022.5}{2023^{31}+5}>1+\dfrac{2022.5}{2023^{32}>5}\)

\(\Leftrightarrow2023A>2023B\Leftrightarrow A>B\)

Ta có :

\(\dfrac{10^{2023}}{10^{2024}}=\dfrac{10^{2022}}{10^{2023}}\)

mà \(\dfrac{10^{2023}}{10^{2024}}>\dfrac{10^{2023}-3}{10^{2024}-3}\)

     \(\dfrac{10^{2022}}{10^{2023}}< \dfrac{10^{2022}+1}{10^{2023}+1}\)

\(\Rightarrow\dfrac{10^{2023}-3}{10^{2024}-3}< \dfrac{10^{2022}+1}{10^{2023}+1}\)

b) \(M=\dfrac{10^{2023}+1}{10^{2024}+1}< 1\) ( Vì tử < mẫu )

Ta có: \(M=\dfrac{10^{2023}+1}{10^{2024}+1}< \dfrac{10^{2023}+1+9}{10^{2024}+1+9}=\dfrac{10^{2023}+10}{10^{2024}+10}=\dfrac{10.\left(10^{2022}+1\right)}{10.\left(10^{2023}+1\right)}=\dfrac{10^{2022}+1}{10^{2023}+1}=N\)

Vì \(\dfrac{10^{2023}+1}{10^{2024}+1}< \dfrac{10^{2022}+1}{10^{2023}+1}\) nên \(M< N\)

Trước hết ta phải chứng minh \(\dfrac{a}{b}< \dfrac{a+1}{b+1}\) (a, b ϵ N; a < b).

Thật vậy, \(\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{a+ab}{b^2+b}\) và \(\dfrac{a+1}{b+1}=\dfrac{\left(a+1\right)b}{\left(b+1\right)b}=\dfrac{ab+b}{b^2+b}\).

Mà theo giả thuyết là a < b nên \(\dfrac{a+ab}{b^2+b}< \dfrac{ab+b}{b^2+b}\), suy ra \(\dfrac{a}{b}< \dfrac{a+1}{b+1}\) (a, b ϵ N; a < b).

Từ đây ta có:

\(B=\dfrac{2022^{2022}+1}{2022^{2023}+1}=\dfrac{2022^{2023}+2022}{2022^{2024}+2022}=\dfrac{2022^{2023}+2021+1}{2022^{2024}+2021+1}\)

Đặt \(A_1=\dfrac{2022^{2023}+2}{2022^{2024}+2}=\dfrac{2022^{2023}+1+1}{2022^{2024}+1+1}\), rõ ràng \(A_1>A\).

Đặt \(A_2=\dfrac{2022^{2023}+3}{2022^{2024}+3}=\dfrac{2022^{2023}+2+1}{2022^{2024}+2+1}\), rõ ràng \(A_2>A_1\).

...

Đặt \(A_{2020}=\dfrac{2022^{2023}+2021}{2022^{2024}+2021}=\dfrac{2022^{2023}+2020+1}{2022^{2024}+2020+1}\), rõ ràng \(A_{2020}>A_{2019}\) và \(B>A_{2020}\).

Suy ra \(B>A_{2020}>A_{2019}>...>A_2>A_1>A\). Vậy A < B.

Ta có A = \(\dfrac{2022^{2023}}{2022^{2024}}=\dfrac{1}{2022}\) ; B = \(\dfrac{2022^{2022}}{2022^{2023}}=\dfrac{1}{2022}\)

Mà \(\dfrac{1}{2022}=\dfrac{1}{2022}\)

Vậy A = B

a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)

b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)

c) \(\dfrac{135}{175}=\dfrac{27}{35}\)

\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)

\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)

\(A=\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\)

\(A=\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)

\(\Rightarrow3A=3.\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)

\(\Rightarrow3A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\)

\(\Rightarrow3A-A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\right)\)

\(\Rightarrow2A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\dfrac{1}{3^1}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...\dfrac{1}{3^{2022}}-\dfrac{1}{3^{2023}}\)

\(\Rightarrow2A=1-\dfrac{1}{3^{2023}}\)

\(\Rightarrow A=\dfrac{1}{2}\left(1-\dfrac{1}{3^{2023}}\right)\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{2023}}< \dfrac{1}{2}\)

\(B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{12}=\dfrac{4+3+1}{12}=\dfrac{8}{12}=\dfrac{2}{3}\)

mà \(\dfrac{2}{3}>\dfrac{1}{2}\) \(\left(\dfrac{2}{3}=\dfrac{4}{6}>\dfrac{1}{2}=\dfrac{3}{6}\right)\)

\(\Rightarrow A< B\)

       A =      \(\dfrac{1}{3}\)+ \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+............+\(\dfrac{1}{3^{2023}}\)

     3A = 1+ \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^{2022}}\)

3A - A =  1 - \(\dfrac{1}{3^{2023}}\)

   2A   = 1 - \(\dfrac{1}{3^{2023}}\) < 1

      B =  \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\)+ \(\dfrac{1}{12}\)

      B  = \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{1}{12}\)

     B   = \(\dfrac{8}{12}\)

     B   = \(\dfrac{2}{3}\) ⇒ 2B = \(\dfrac{4}{3}\) > 1 

2A < 2B ⇒ A < B