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 \(M< N\)

...

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

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

Ta có

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

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

\(\Rightarrow C>D\)

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 : \(M=2023.2023=2023^2\)

\(\Rightarrow M>N\)

Ta có :

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

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

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

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

Sửa đề: \(a=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)

Ta có: \(a=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)

=>\(3a=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2023}{3^{2022}}-\frac{2024}{3^{2023}}\)

=>\(3a+a=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2023}{3^{2022}}-\frac{2024}{3^{2023}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)

=>\(4a=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}-\frac{2024}{3^{2024}}\)

Đặt \(b=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}\)

=>\(3b=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2022}}\)

=>\(3b+b=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2022}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}\)

=>\(4b=-1-\frac{1}{3^{2023}}=\frac{-3^{2023}-1}{3^{2023}}\)

=>\(b=\frac{-3^{2023}-1}{4\cdot3^{2023}}\)

Ta có: \(4a=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}-\frac{2024}{3^{2024}}\)

=>\(4a=1+\frac{-3^{2023}-1}{4\cdot3^{2023}}-\frac{2024}{3^{2024}}=1+\frac{-3^{2024}-3}{4\cdot3^{2024}}-\frac{8096}{4\cdot3^{2024}}\)

=>\(4a=1-\frac{3^{2024}+8099}{4\cdot3^{2024}}=1-\frac14-\frac{8099}{4\cdot3^{2024}}=\frac34-\frac{8099}{4\cdot3^{2024}}\)

=>\(4a<\frac34\)

=>\(a<\frac{3}{16}\)

mà \(\frac{3}{16}<1<\frac{20}{3}\)

nên \(a<\frac{20}{3}\)