1)

-2/15 < 0 

-10/-11 > 0

nên x < y

2) 

ví dụ 1 < 2 => -1 > -2

ta có 16/29 < 16/27 < 19/27 

suy ra -16/29 > -16/27 > -19/27

Giải:

a) \(\dfrac{-3}{7}\) và \(\dfrac{2}{-5}\) 

\(\dfrac{-3}{7}=\dfrac{-3.5}{7.5}=\dfrac{-15}{35}\) 

\(\dfrac{2}{-5}=\dfrac{-2}{5}=\dfrac{-2.7}{5.7}=\dfrac{-14}{35}\) 

Vì \(\dfrac{-15}{35}< \dfrac{-14}{35}\) nên \(\dfrac{-3}{7}< \dfrac{2}{-5}\) 

b) \(\dfrac{-6}{7}\) và \(\dfrac{19}{23}\) 

Vì \(\dfrac{-6}{7}\) là số âm mà \(\dfrac{19}{23}\) là số dương nên \(\dfrac{-6}{7}< \dfrac{19}{23}\) 

c) \(\dfrac{-11}{23}\) và \(\dfrac{-13}{21}\) 

\(\dfrac{-11}{23}=\dfrac{-11.13}{23.13}=\dfrac{-143}{299}\) 

\(\dfrac{-13}{21}=\dfrac{-13.11}{21.11}=\dfrac{-143}{231}\) 

Vì \(\dfrac{-143}{299}>\dfrac{-143}{231}\) nên \(\dfrac{-11}{23}>\dfrac{-13}{21}\) 

d) \(\dfrac{-1}{5}\) và \(\dfrac{1}{100}\) 

Vì \(\dfrac{-1}{5}\) là số âm mà \(\dfrac{1}{100}\) là số dương nên \(\dfrac{-1}{5}< \dfrac{1}{100}\) 

Giải: (tiếp)

e) \(\dfrac{267}{-268}\) và \(\dfrac{-1347}{1343}\) 

\(\dfrac{267}{-268}=\dfrac{-267}{268}=\dfrac{-267.449}{268.449}=\dfrac{-119883}{120332}\) 

\(\dfrac{-1347}{1343}=\dfrac{-1347.89}{1343.89}=\dfrac{-119883}{119527}\)  

Vì \(\dfrac{-119883}{120332}>\dfrac{-119883}{119527}\) nên \(\dfrac{267}{-268}>\dfrac{-1347}{1343}\)

f) \(\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\) 

\(\dfrac{-13}{38}=\dfrac{-13.29}{38.29}=\dfrac{-377}{1102}\)  

\(\dfrac{29}{-88}=\dfrac{-29}{88}=\dfrac{-29.13}{88.13}=\dfrac{-377}{1144}\) 

Vì \(\dfrac{-377}{1102}< \dfrac{-377}{1144}\) nên \(\dfrac{-13}{38}< \dfrac{29}{-88}\) 

g) \(\dfrac{4}{9}\) và \(\dfrac{13}{18}\) 

\(\dfrac{4}{9}=\dfrac{4.2}{9.2}=\dfrac{8}{18}\) 

Vì \(\dfrac{8}{18}< \dfrac{13}{18}\) nên \(\dfrac{4}{9}< \dfrac{13}{18}\) 

h) \(\dfrac{-15}{7}\) và \(\dfrac{-6}{5}\) 

\(\dfrac{-15}{7}=\dfrac{-15.5}{7.5}=\dfrac{-75}{35}\) 

\(\dfrac{-6}{5}=\dfrac{-6.7}{5.7}=\dfrac{-42}{35}\) 

Vì \(\dfrac{-75}{35}< \dfrac{-42}{35}\) nên \(\dfrac{-15}{7}< \dfrac{-6}{5}\)

a: \(x_1=x_2=x_3=x_4\)

(Sửa \(cn-bm\rightarrow cn-dm\))

Ta có :

\(\left\{{}\begin{matrix}ad-bc=1\\cn-dm=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}ad=1+bc\\cn=1+dm\end{matrix}\right.\)

\(\dfrac{x}{y}=\dfrac{a}{b}.\dfrac{d}{c}=\dfrac{ad}{bc}=\dfrac{1+bc}{bc}=1+\dfrac{1}{bc}>1\left(bc>0\right)\)

\(\Rightarrow x=\dfrac{a}{b}>y=\dfrac{c}{d}\left(2\right)\)

\(\dfrac{y}{z}=\dfrac{c}{d}.\dfrac{n}{m}=\dfrac{cn}{dm}=\dfrac{1+dm}{dm}=1+\dfrac{1}{dm}>1\left(dc>0\right)\)

\(\Rightarrow y=\dfrac{c}{d}>z=\dfrac{m}{n}\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow x>y>z\)

a, \(\frac{15}{106}\)và \(\frac{21}{133}\)

          Ta có:

\(\frac{15}{106}< \frac{15}{100}=\frac{3}{20}=\frac{21}{140}< \frac{21}{133}\)

\(\Rightarrow\frac{15}{106}< \frac{21}{133}\)

             Vậy ........

b, \(\frac{31}{100}\)và \(\frac{89}{150}\)

       Ta có:

\(\frac{31}{100}< \frac{31}{93}=\frac{1}{3}=\frac{50}{150}< \frac{89}{150}\)

\(\Rightarrow\frac{31}{100}< \frac{89}{150}\)

        Vậy........

c, \(\frac{2020}{2019}\)và \(\frac{2021}{2020}\)

           Ta có:

\(\frac{2020}{2019}-1=\frac{1}{2019}\)     ;

\(\frac{2021}{2020}-1=\frac{1}{2020}\)

    Vì \(\frac{1}{2019}>\frac{1}{2020}\)

               \(\Rightarrow\frac{2020}{2019}-1>\frac{2021}{2020}-1\)  

              \(\Rightarrow\frac{2020}{2019}>\frac{2021}{2020}\)

 Vậy .........

d, n+2019/n+2021 và n+2020/n+2022

Câu d bn tự lm nhé

Cảm ơn bạn nhiều lắm! THANK YOU VERY MUCH!!!!!!!!!

\(\frac{127}{-128}=\frac{-127}{128}\Rightarrow\frac{-128}{128}=-1\left(1\right).\)

\(\frac{-1345}{1344}=\frac{1344}{1344}=-1\left(2\right)\)

Từ (1) và (2) ta suy ra :

\(\Rightarrow\frac{-1345}{1344}=-1=\frac{127}{-128}\)

\(\Rightarrow\frac{-1345}{1344}=\frac{127}{-128}\)

\(\frac{-37}{946}< \frac{-1}{8}\)

giải thick hộ mình vì so lại bé hơn

106/204>107/206

 so sánh bằng phân số trung gian

vì \(\frac{106}{204}\)> \(\frac{106}{206}\)nên \(\frac{106}{204}\)> \(\frac{107}{206}\)