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

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

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

Bài 1:

Ta thấy A < 1

=> A = \(\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}=\frac{17^{18}+17}{17^{19}+17}=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=\frac{17^{17}+1}{17^{18}+1}=B\)

Vậy A < B

Bài 2:

Ta thấy C < 1

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

Vậy C < D

\(\frac{9^{99}-1}{-9^{98}+1}\) < \(\frac{-9^{98}-1}{9^{97+1}}\)

làm on trình bày cách giải

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

CÁCH 1

Ta có \(A=\frac{89}{99}=\frac{99-1}{99}=\frac{99}{99}-\frac{1}{99}=1-\frac{1}{99}\)

\(B=\frac{98.99+1}{98.99}=\frac{98.99}{98.99}+\frac{1}{98.99}\)

Vì \(\frac{1}{98.99}< \frac{1}{99}\Rightarrow1+\frac{1}{98.99}>1-\frac{1}{99}\Rightarrow\frac{98.99+1}{98.99}>\frac{98}{99}\Rightarrow B>A\)

CÁCH 2 

Ta thấy 98 < 99 nên \(\frac{98}{99}< 1\)hay \(A< 1\)

Ta thấy \(98.99+1>98.99\Rightarrow\frac{98.99}{98.99+1}>1\Rightarrow B>1\)

Vì A < 1 ; B > 1 nên A < B

\(A=\frac{98}{99}< 1;\Rightarrow A< 1\)

\(B=\frac{98.99+1}{98.99}\)

Ta loại các số chia hết cho nhau thì được

\(B=\frac{1.1+1}{1.1}=1+1=2\)

\(2>1;\Rightarrow B>1;\Rightarrow B>A\)

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