99/100<1

100/99>1

=>99/100<100/99

Ta có: 99/100 < 1  
          100/99 > 1
=> 99/100 < 100/99

so sánh với 1. 99/100 < 1 và 100/99 > 1 nên 99/100 < 100/99

\(\dfrac{97}{100}\)  và \(\dfrac{98}{99}\)

\(\dfrac{97}{100}=\dfrac{97\times99}{100\times99}=\dfrac{9603}{9900}\)

\(\dfrac{98}{99}=\dfrac{98\times100}{99\times100}=\dfrac{9800}{9900}\)

Vì: \(9603< 9800\)  nên => \(\dfrac{97}{100}< \dfrac{98}{99}\)

\(\dfrac{13}{17}\)  và \(\dfrac{131}{171}\)

\(\dfrac{13}{17}=\dfrac{13\times171}{17\times171}=\dfrac{2223}{2907}\)

\(\dfrac{131}{171}=\dfrac{131\times17}{171\times17}=\dfrac{2227}{2907}\)

Vì: \(2227>2223\)  nên: => \(\dfrac{13}{17}< \dfrac{131}{171}\)

\(\dfrac{51}{61}\)  và \(\dfrac{515}{616}\)

\(\dfrac{51}{61}=\dfrac{51\times616}{61\times616}=\dfrac{31416}{37576}\)

\(\dfrac{515}{616}=\dfrac{515\times61}{616\times61}=\dfrac{31415}{37576}\)

Vì: \(31416>31415\)  Nên => \(\dfrac{51}{61}>\dfrac{515}{616}\)

a/

$\frac{97}{100}< \frac{98}{100}< \frac{98}{99}$

c/

$\frac{131}{171}=1-\frac{40}{171}> 1-\frac{40}{170}=1-\frac{4}{17}=\frac{13}{17}$
d/

$\frac{51}{61}=1-\frac{10}{61}=1-\frac{100}{610}$

$\frac{515}{616}=1-\frac{101}{616}$

Xét hiệu:

$\frac{100}{610}-\frac{101}{616}=\frac{100.616-101.610}{610.616}$

$=\frac{100(610+6)-101.610}{610.616}$

$=\frac{600-610}{610.616}<0$

$\Rightarrow \frac{100}{610}< \frac{101}{616}$

$\Rightarrow 1-\frac{100}{610}> 1-\frac{101}{616}$

$\Rightarrow \frac{51}{61}> \frac{515}{616}$ 

gọi A là phân số thứ nhất, B là phân số thứ 2

\(\frac{A}{3}=\frac{3^{100}+1}{3^{100}+3}\)có phần bù là \(\frac{2}{3^{100}+3}\)

\(\frac{B}{3}=\frac{3^{99}+1}{3^{99}+3}\)có phần bù là \(\frac{2}{3^{99}+3}\)

ta thấy \(\frac{2}{3^{100}+3}< \frac{2}{3^{99}+3}\Rightarrow A>B\)

mink nghĩ vậy bạn ạ

a) \(\dfrac{122}{123}\) và \(\dfrac{10}{11}\)

\(1-\dfrac{122}{123}=\dfrac{1}{123}\)              

\(1-\dfrac{10}{11}=\dfrac{1}{11}\)

Vì \(\dfrac{1}{123}< \dfrac{1}{11}\) nên ⇒ \(\dfrac{122}{123}< \dfrac{10}{11}\) 

b) \(\dfrac{16}{12}\) và \(\dfrac{99}{100}\)

\(\dfrac{16}{12}>1\) và \(\dfrac{99}{100}< 1\)

⇒ \(\dfrac{16}{12}>\dfrac{99}{100}\)

c) \(\dfrac{35}{70}\) và \(\dfrac{6}{11}\)

\(\dfrac{35}{70}=\dfrac{1}{2}\) = \(\dfrac{6}{12}\)

Vì \(\dfrac{6}{12}< \dfrac{6}{11}\) nên ⇒ \(\dfrac{35}{70}< \dfrac{6}{11}\)

a) \(\frac{8}{9}=1-\frac{1}{9}\)  

\(\frac{108}{109}=1-\frac{1}{109}\)  

Vì \(\frac{1}{9}>\frac{1}{109}\)  

Nên \(1-\frac{1}{9}< 1-\frac{1}{109}\)   

Vậy \(\frac{8}{9}< \frac{108}{109}\)  

b) 

\(\frac{97}{100}=\frac{97\cdot99}{100\cdot99}\)  

\(\frac{98}{99}=\frac{98\cdot100}{99\cdot100}\) 

\(\Rightarrow\frac{97}{100}< \frac{98}{99}\)

\(a,\frac{8}{9}< \frac{108}{109}\)

\(b,\frac{97}{100}< \frac{98}{99}\)

\(c,\frac{19}{18}>\frac{2017}{2016}\)

\(d,\frac{15}{16}>\frac{515}{616}\)