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

\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+17}{17^{19}+17}\)

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

=> A < B

(98^99-1)/(98^98-1)

1) Phân tích A ra :

 A= 17¹⁷.17+\(\frac{1}{17^{18}.17}\)+1 So sánh với B ta có: A có 17¹⁸>17¹⁷ của B nhưng B lại có 1/17¹⁸>1/17¹⁹.

Mà 17¹⁸>1/17¹⁸ nên suy ra A>B

2) Bài nay tương tự bài trên. 

2/(2012+2013) < 2/(2012 + 2012) = 2/ (2.2012) = 1/2012 
2009/(2012+2013) < 2009/2012 

=> 2011/(2012+2013) = 2/(2012+2013) + 2009/(2012+2013) < 1/2012 + 2009/2012 
=> 2011/(2012+2013) < 2010/2012 (a) 

2012/(2012+2013) < 2012/2013 (b) 

lấy (a) + (b) => (2011+2012)/(2012+2013) < 2010/2012 + 2012/2013 

vậy B < A 

trung nguyên làm bài j vậy

Bài 1: 

1: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)

\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)

mà \(17^{19}+1>17^{18}+1\)

nên 17A>17B

hay A>B

2: \(C=\dfrac{98^{99}+98^{10}+1-98^{10}}{98^{89}+1}=98^{10}+\dfrac{1-98^{10}}{98^{89}+1}\)

\(D=\dfrac{98^{98}+98^{10}+1-98^{10}}{98^{88}+1}=98^{10}+\dfrac{1-98^{10}}{98^{88}+1}\)

mà \(98^{89}+1>98^{88}+1\)

nên C>D

a, \(A=\frac{17^{18}+1}{17^{19}+1}< 1\)

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

\(\Rightarrow A< B\)

b, Tương tự câu a

a)Ta có : 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

b) Làm tương tự như câu A

ta co B=17¹⁷+1/17¹⁸+1=10x(17¹⁷+1)/10x(17¹⁸+1)

                                 =17¹⁸+10/17¹⁹+10

mà A=17¹⁸+1/17¹⁹+1 < B=17¹⁸+10/17¹⁹+10

suy ra A < B

C=98^99+1/98^89+1 > 98^99+1+97/98^89+1+97=98^99+98/98^89+98=98(98^98+1/98^88+1)=D

Vậy C>D

Bài 6.7*

Ta có : \(\dfrac{17^{18}+1}{17^{19}+1}< 1\)

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

\(\)Vậy A < B

Bài 6.6*

Ta có : \(\dfrac{98^{99}+1}{98^{89}+1}>1\)

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

Vậy 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