Ta có \(\dfrac{2004}{2005}=\dfrac{2005-1}{2005}=1-\dfrac{1}{2005}\)

       \(\dfrac{2005}{2006}=\dfrac{2006-1}{2006}=1-\dfrac{1}{2006}\)

Vì 2005<2006=>\(\dfrac{1}{2005}>\dfrac{1}{2006}\)

                      =>\(1-\dfrac{1}{2005}< 1-\dfrac{1}{2006}\)

                      \(\Leftrightarrow\dfrac{2004}{2005}< \dfrac{2005}{2006}\)

= 

2004/2005=1-1/2005

2005/2006=1-1/2006

mà 1/2005>1/2006

nên 2004/2005<2005/2006

a) \(\dfrac{4}{9}< \dfrac{4}{8}=\dfrac{1}{2}\)

b) \(\dfrac{5}{8}=\dfrac{15}{24}>\dfrac{14}{24}=\dfrac{7}{12}\)

a)
4/9 < 5/10 mà 5/10 =1/2
=> 4/9<1/2
 

a: \(\dfrac{4}{9}< \dfrac{1}{2}\)

b: \(\dfrac{5}{8}>\dfrac{7}{12}\)

c: \(-\dfrac{5}{8}>-\dfrac{17}{18}\)

d: \(-\dfrac{8}{15}>-\dfrac{2}{3}\)

a: 31/32>0>-5/57

b: -15/81<0<7/90

3/

a/ \(\dfrac{31}{32}>0>\dfrac{-5}{57}\)

b/ \(\dfrac{-15}{81}< 0< \dfrac{7}{90}\)

\(\dfrac{31}{32}>0>\dfrac{-5}{57}\)

Ta có \(\dfrac{{31}}{{32}} > 0\) và \(\dfrac{{ - 5}}{{57}} < 0\) nên \(\dfrac{{31}}{{32}} > \dfrac{{ - 5}}{{57}}\).

a) \(\dfrac{2}{5}=\dfrac{4}{10}\)

\(\dfrac{4}{10}>\dfrac{3}{10}\)

b) \(\dfrac{5}{6}=\dfrac{10}{12}\)

\(\dfrac{7}{12}< \dfrac{10}{12}\)

c) \(\dfrac{1}{2}=\dfrac{2}{4}\)

\(\dfrac{3}{4}< \dfrac{2}{4}\)

d) \(\dfrac{8}{3}=\dfrac{56}{21}\)

\(\dfrac{56}{21}>\dfrac{11}{21}\)

Câu b thì gg search nhé

a: 

13/17=1-4/17

8/12=1-4/12

mà 4/17<4/12

nên 13/17>8/12=12/18

b: 16/51<17/51=1/3=30/90<31/90

12/18<13/17

16/51>31/90

a) \(\dfrac{3}{4}=\dfrac{3\times4}{4\times4}=\dfrac{12}{16}\)

b) \(\dfrac{1}{3}=\dfrac{1\times3}{3\times3}=\dfrac{3}{9}\)

c) \(\dfrac{5}{6}=\dfrac{5\times3}{6\times3}=\dfrac{15}{18}\)

Trả lời:

2/5 : 4/13 = (2x13); (5x4)= 26:20 >1

=> 2/5> 4/13

2 / 5 > 4 / 13

Câu này nếu ko quy đồng mẫu số hoặc tử số thì không tính được