a, ta có :

 \(\hept{\begin{cases}\frac{6}{7}< 1\\\frac{11}{10}>1\end{cases}}\Rightarrow\frac{6}{7}< \frac{11}{10}\)

b, ta có :

\(\hept{\begin{cases}\frac{-5}{7}< 0\\\frac{2}{7}>0\end{cases}}\Rightarrow\frac{-5}{7}>\frac{2}{7}\)

c, ta có :

\(\hept{\begin{cases}\frac{419}{-723}< 0\\\frac{-697}{-313}>0\end{cases}}\Rightarrow\frac{419}{-723}< \frac{-679}{-313}\)

a]6/7 va 11/10 

60/70 và 77/70 vay 11/10 lớn hơn 6/7 

a) \(\frac{6}{7}\) và \(\frac{11}{10}\)

Vì \(\frac{6}{7}< 1;1< \frac{11}{10}\)

\(\Rightarrow\frac{6}{7}< \frac{11}{10}\)

Vậy \(\frac{6}{7}< \frac{11}{10}\)

1. ​​a, \(\frac{6}{7}\)=\(\frac{60}{70}\);\(\frac{11}{10}\)=\(\frac{77}{70}\)

vì \(\frac{60}{70}\)<\(\frac{77}{70}\)nên \(\frac{6}{7}\)<\(\frac{11}{10}\)

b, \(\frac{-5}{17}\)<0<\(\frac{2}{7}\)

c, \(\frac{419}{-723}\)<0<\(\frac{-697}{-313}\)

2.

Ta có :\(\frac{2}{6}\)=\(\frac{20}{60}\);\(\frac{5}{12}\)=\(\frac{25}{60}\);\(\frac{4}{15}\)=\(\frac{16}{60}\);\(\frac{8}{20}\)=\(\frac{24}{60}\);\(\frac{10}{30}\)=\(\frac{20}{60}\)

Vì \(\frac{16}{60}\)<\(\frac{20}{60}\)<\(\frac{24}{60}\)<\(\frac{25}{60}\)nên \(\frac{4}{15}\)<\(\frac{2}{6}\)=\(\frac{10}{30}\)<\(\frac{8}{20}\)<\(\frac{5}{12}\)

6/7<1 và 11/10>1 => 6/7<11/10

(-5)/17<0 và 2/7>0 => (-5)/17<2/7

419/(-723)<0 và -697/-313>0 => 419/-723< -697/-313

a. \(\frac{6}{7}=\frac{60}{70};\frac{11}{10}=\frac{77}{70}\)

\(\frac{77}{70}>\frac{60}{70}\rightarrow\frac{11}{10}>\frac{6}{7}\)

b. \(\frac{-5}{17}< 0;\frac{2}{7}>0\rightarrow\frac{2}{7}>\frac{-5}{17}\)

c. \(\frac{419}{-723}< 0;\frac{-697}{-313}=\frac{697}{313}>0\Rightarrow\frac{-697}{-313}>\frac{419}{-723}\)

a)Ta có : 6/7 < 1

11/10 >1

Suy ra : 6/7 < 11/10

B) -5/17 < 0 và 2/7 >0

Suy ra -5/17 < 2/7

c) câu c mk ko bt xl bn nha

a)  16 − 19 < 0 < 15 17

b)  419 − 723 < 0 < − 697 − 313

6/7=60/70

11/10=77/70

Vì 60/70< 77/70 nên 6/7<11/10