3457/3452 < 6789/6784

\(\frac{3457}{3452}=\frac{3452+5}{3452}=1+\frac{5}{3452},\frac{6789}{6784}=\frac{6784+5}{6789}=1+\frac{5}{6789}\)

Ta có: \(3452< 6789\Leftrightarrow\frac{1}{3452}>\frac{1}{6789}\Leftrightarrow\frac{5}{3452}>\frac{5}{6789}\)

suy ra \(\frac{3457}{3452}>\frac{6789}{6784}\).

Trả lời :

Ta có:

3457/3452=1+ (5/3452)

6789/6784=1+(5/6784)

Vì 1=1 nên ta so sánh 5/3452 và 5/6784

Vì 5=5 , 3452<6784

Nên 5/3452>5/6784

Do đó: 1+(5/3452)>1+(5/6784)

hay 3457/3452>6789/6784

3457/3452=1+5/3452

6789/6784=1+5/6784

mà 3452>6784

nên 3457/3452<6789/6784

>

>

>

a. 19/10 > 10/11

b. 11/10 = 12/11

c. 9/10 = 10/11

a)\(\dfrac{19}{10}>\dfrac{10}{11}\)

b)\(\dfrac{11}{10}=\dfrac{12}{11}\)

c)\(\dfrac{9}{10}< \dfrac{10}{11}\)

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

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

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

Câu b thì gg search nhé

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

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

3/

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

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

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