sắp xêp theo thứ tự tăng dần là: -3/4 ; -4/8;-1/3;1/2;5/2

Sắp xếp theo thứ tự từ bé đến lớn 

9/5;1/6;15/7

\(\frac{1}{6},\frac{1}{3}\frac{3}{2},\frac{5}{2}\)

1/6 ; 1/3 ; 3/2 ; 5/2 .

\(\dfrac{3}{4}=\dfrac{3\times5}{4\times5}=\dfrac{15}{20}\)

\(\dfrac{2}{4}=\dfrac{2\times5}{4\times5}=\dfrac{10}{20}\)

\(1=\dfrac{20}{20}\)

\(\dfrac{6}{5}=\dfrac{6\times4}{5\times4}=\dfrac{24}{20}\)

Vì \(\dfrac{10}{20}< \dfrac{15}{20}< \dfrac{20}{20}< \dfrac{24}{20}\) nên \(\dfrac{2}{4}< \dfrac{3}{4}< 1< \dfrac{6}{5}\)

1. \(\frac{7}{39}=\frac{140}{780};\frac{11}{65}=\frac{132}{780};\frac{9}{52}=\frac{135}{780}\) và thứ tự tăng dần là : \(\frac{132}{780}< \frac{135}{780}< \frac{140}{780}\)

2. \(\frac{17}{20}=\frac{153}{180};-\frac{19}{30}=-\frac{114}{180};\frac{38}{45}=\frac{152}{180};-\frac{13}{10}=-\frac{130}{180}\) và thứ tự tăng dận là : 

\(-\frac{130}{180}< -\frac{114}{180}< \frac{152}{180}< \frac{153}{180}\)

.

a) -12,13; -2,4; -2,3; 0,5; 2,4

b) 2,999; 2,9; -2,9; -2,999

a) -12,13; -2,4; -2,3; 0,5; 2,4

b) 2,999; 2,9; -2,9; -2,999

a) -12,13; -2,4; -2,3; 0,5; 2,4

b) 2,999; 2,9; -2,9; -2,999

Ta có: \(\frac{-3}{5}=\frac{-3\cdot16}{5\cdot16}=\frac{-48}{80};\frac{4}{-8}=\frac{-4}{8}=\frac{-4\cdot10}{8\cdot10}=\frac{-40}{80}\)

\(\frac54=\frac{5\cdot20}{4\cdot20}=\frac{100}{80};\frac{3}{-10}=\frac{-3}{10}=\frac{-3\cdot8}{10\cdot8}=-\frac{24}{80}\) ;

\(\frac{-12}{30}=\frac{-12:6}{30:6}=\frac{-2}{5}=\frac{-2\cdot16}{5\cdot16}=\frac{-32}{80}\)

\(\frac32=\frac{3\cdot40}{2\cdot40}=\frac{120}{80}\)

mà \(-\frac{48}{80}<-\frac{40}{80}<-\frac{32}{80}<-\frac{24}{80}<\frac{100}{80}<\frac{120}{80}\)

nên \(-\frac35<-\frac12<-\frac25<\frac{3}{-10}<\frac54<\frac32\)

lớn hơn 1

Mà (vì 6 > 3) và (3 < 5)

nên ta có kết luận là:

Ta có: \(-\frac{12}{7}<\frac{-7}{7}=-1\)

\(-1=\frac{-40}{40}<\frac{-35}{40}<\frac{-24}{40}\)

=>\(-1<-\frac78<-\frac35\)

Do đó: \(-\frac{12}{7}<-1<-\frac78<-\frac35<0\) (1)

Ta có: \(\frac25=\frac{2\cdot6}{5\cdot6}=\frac{12}{30};\frac{13}{5}=\frac{13\cdot6}{5\cdot6}=\frac{78}{30}\)

\(\frac{7}{15}=\frac{7\cdot2}{15\cdot2}=\frac{14}{30};1=\frac{1\cdot30}{30}=\frac{30}{30};\frac92=\frac{9\cdot15}{2\cdot15}=\frac{135}{30}\)

mà \(\frac{12}{30}<\frac{14}{30}<\frac{30}{30}<\frac{78}{30}<\frac{135}{30}\)

nên \(0<\frac25<\frac{7}{15}<1<\frac{13}{5}<\frac92\) (2)

Từ (1),(2) suy ra \(-\frac{12}{7}<-1<-\frac78<-\frac35<0<\frac25<\frac{7}{15}<1<\frac{13}{5}<\frac92\)