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

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

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

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

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

\(\Rightarrow C>D\)

C < D

Chắc chắn

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

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

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

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

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

Vì C>1 còn D=1 nên C>D

dung cho mih nha

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

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

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

Giúp tớ với ngày kia tớ kiểm tra rồi

Ai đúng mk cho t**k nha! Thanks

#Bạn gì ơi,bài này có giáo viên làm rồi nhé,link đây: https://olm.vn/hoi-dap/detail/36594660334.html

#C > D