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

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

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

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

\(\Rightarrow C>D\)

Lấy C - D

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

Tử số bằng:

\(98^{187}+98^{99}+98^{88}+1-98^{187}-98^{98}-98^{89}-1\)

=\(98^{99}+98^{88}-98^{98}-98^{89}\)

= \(98^{99}-98^{98}+98^{88}-98^{89}\)

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

= \(98^{98}.97-98^{88}.97=97\left(98^{98}-98^{88}\right)>0\)

Vậy C - D > 0 => C > D

Do C>1 nên ta có:

C=98⁹⁹+1/98⁸⁹+1>98⁹⁹+1+97/98⁸⁹+1+97=98⁹⁹+98/98⁸⁹+98=98(98⁹⁸+1)/98(98⁸⁸+1)=98⁹⁸+1/98⁸⁸+1=D

suy ra C>D

do c lon hon 1 nen ta co:

c=98⁹⁹+1/98⁸⁹+1>98⁹⁹+1+97/98⁸⁹+1+97=98⁹⁹+98/98⁸⁹+98=98(98⁹⁸+1)/98(98⁹⁸+1)=98⁹⁸+1/98⁸⁸+1=D

tu do tu se suy ra duoc la D>C do

\(A=\frac{-\left(98^{98}+1\right)}{-\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}\)

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

A-1=\(\frac{98^{98}-98^{88}}{98^{88}+1}=\frac{98^{88}.\left(98^{10}-1\right)}{98^{88}+1}\)

B-1=\(\frac{98^{99}-98^{89}}{98^{89}+1}=\frac{98^{89}.\left(98^{10}-1\right)}{98^{89}+1}\)

=>\(\frac{A-1}{B-1}=\frac{98^{88}.\left(98^{10}-1\right)}{98^{88}+1}.\frac{98^{89}+1}{98^{89}.\left(98^{10}-1\right)}=\frac{98^{89}+1}{98.\left(98^{88}+1\right)}=\frac{98^{89}+1}{98^{89}+98}< 1\)

->A-1<B-1

->A<B

C < D

Chắc chắn

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

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

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

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