\(C-D=\dfrac{\left(98^{99}+1\right)\left(98^{88}+1\right)-\left(98^{89}+1\right)\left(98^{98}+1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)

\(=\dfrac{98^{187}+98^{99}+98^{88}+1-98^{197}-98^{89}-98^{98}-1}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)

\(=\dfrac{98^{99}-98^{98}+98^{88}-98^{89}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{98^{98}\left(98-1\right)-98^{88}\left(98-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)

\(=\dfrac{97.98^{98}-97.98^{88}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{97.98^{88}\left(98^{10}-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}>0\)

\(\Rightarrow C>D\)

C < D

Chắc chắn

\(C=\frac{98^{99}+1}{98^{89}+1}\)

\(D=\frac{98^{98}+1}{98^{88}+1}\)

\(C< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98^{98}\left(98+1\right)}{98^{88}\left(98+1\right)}\)

\(C< \frac{98^{98}}{98^{88}}=D\)

mil đang phân vân cả 3 đáp án : >,<,=

C > D

a: \(98^{10}\cdot A=\dfrac{98^{98}+98^{10}}{98^{98}+1}=1+\dfrac{98^{10}-1}{98^{98}+1}\)

\(98^{10}\cdot B=\dfrac{98^{99}+98^{10}}{98^{99}+1}=1+\dfrac{98^{10}-1}{98^{99}+1}\)

98^88+1>98^99+1

=>A<B

b: \(\dfrac{1}{2022^2}\cdot C=\dfrac{2022^{2023}+1}{2022^{2023}+2022^2}=1+\dfrac{1-2022^2}{2022^{2023}+2022^2}\)

\(\dfrac{1}{2022^2}\cdot D=\dfrac{2022^{2021}+1}{2022^{2021}+2022^2}=1+\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

2022^2023>2022^2021

=>2022^2023+2022^2>2022^2021+2022^2

=>\(\dfrac{2022^2-1}{2022^{2023}+2022^2}< \dfrac{2022^2-1}{2022^{2021}+2022^2}\)

=>\(\dfrac{1-2022^2}{2022^{2023}+2022^2}>\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

=>C>D

\(C=\frac{98^{99}+1}{98^{88}+1}\)\(D=\frac{98^{98}+1}{98^{98}+1}\)

Vì C>1 còn D=1 nên C>D

dung cho mih nha

Ta có:C=\(\frac{98^{99}+1}{98^{89}+1}\Rightarrow\frac{98^{99}+1}{98^{99}+10}=\frac{98^{99}+1}{98^{99}+1+9}=\frac{98^{99}+1}{1+9}\)

D\(\frac{98^{98}+1}{98^{88}+1}=\frac{98^{98}+1}{98^{98}+10}=\frac{98^{98}+1}{98^{98}+1+9}\frac{98^{98}+1}{1+9}\)

Vì\(\frac{98^{99}+1}{1+9}\)>\(\frac{98^{98}+1}{1+9}\)

=>C>D