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Vì C= \(\dfrac{98^{99}+1}{98^{89}+1}\)>1 thì nên áp dụng tính chất . Nên \(\dfrac{a}{b}\)>1 thì \(\dfrac{a}{b}\)>\(\dfrac{a+m}{b+m}\) ( a∈ N , b và m ∈ N✳) Ta có : C= \(\dfrac{98^{99}+1}{98^{89}+1}\)> \(\dfrac{98^{99}+1+97}{98^{89}+1+97}\)= \(\dfrac{98^{99}+98}{98^{89}+98}\) = \(\dfrac{98.98^{98}+98.1}{98.98^{88}+98.1}\) = \(\dfrac{98.\left(98^{98}+1\right)}{98.\left(98^{88}+1\right)}\)= \(\dfrac{98^{98}+1}{98^{88}+1}\)= B ⇔ Vậy \(\dfrac{98^{99}+1}{98^{89}+1}\)< \(\dfrac{98^{89}+1}{98^{88}+1}\) nên C<D
98 <1
99
98.99+1 Vì 98.99+1 >98.99 nên 98.99+1 >1
98.99 98.99
Suy ra: 98 < 98.99+1
99 98.99
A= \(\frac{98}{99}\)< \(1\)
B= \(\frac{98.99+1}{98.99}\)=\(\frac{98.99}{98.99}+\frac{1}{98.99}\)=\(1+\frac{1}{98.99}\)> 1
=> A<1<B => A<B
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
\(A=\frac{-\left(98^{98}+1\right)}{-\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}\)
\(B=\frac{98^{99}+1}{98^{89}+1}\)
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
\(C-D=\dfrac{\left(98^{99}+1\right)\left(98^{88}+1\right)-\left(98^{89}+1\right)\left(98^{98}+1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{98^{187}+98^{99}+98^{88}+1-98^{197}-98^{89}-98^{98}-1}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{98^{99}-98^{98}+98^{88}-98^{89}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{98^{98}\left(98-1\right)-98^{88}\left(98-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{97.98^{98}-97.98^{88}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{97.98^{88}\left(98^{10}-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}>0\)
\(\Rightarrow C>D\)