Bài 1:

1)mik ko biết trục số ở đâu nên tham khảo:

2

-0,75 <5/3

a: \(x_1=x_2=x_3=x_4\)

(Sửa \(cn-bm\rightarrow cn-dm\))

Ta có :

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

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

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

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

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

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

Lời giải:
a. $\frac{3}{-7}=\frac{-27}{63}$

$\frac{-5}{9}=\frac{-35}{63}$

Do $\frac{27}{63}< \frac{35}{63}$ nên $\frac{-27}{63}> \frac{-35}{63}$

$\Rightarrow \frac{3}{-7}> \frac{-5}{9}$

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b.

$-0,625=\frac{-625}{1000}=\frac{-5}{8}=\frac{-125}{200}$

$\frac{-19}{50}=\frac{-76}{200}> \frac{-125}{200}$

$\Rightarrow -0,625> \frac{-19}{50}$

c.

$-2\frac{5}{9}=-(2+\frac{5}{9})=\frac{-23}{9}=-(\frac{-23}{-9})$

a: -3/100=-9/300; -2/3=-200/300

=>-3/100>-2/3

b: -3/5=-9/15

-2/3=-10/15

=>-3/5>-2/3

c: -5/4<-1<-3/8

d: -2/3=-8/12; -3/4=-9/12

=>-2/3>-3/4

e: -267/268>-1

-1>-1347/1343

=>-267/268>-1347/1343

Câu 1 : 

\(\dfrac{-25}{37}\&\dfrac{-20}{31}\)

Ta thấy \(\dfrac{-25}{37}< \dfrac{-20}{37}\)

mà \(\dfrac{-20}{37}< \dfrac{-20}{31}\)

\(\Rightarrow\dfrac{-25}{37}< \dfrac{-20}{31}\)

Câu 2 :

\(\dfrac{2}{3}\&\dfrac{5}{7}\)

\(\dfrac{2}{3}:\dfrac{5}{7}=\dfrac{2}{3}.\dfrac{7}{5}=\dfrac{14}{15}< 1\)

Ta thấy \(\dfrac{8}{13}:\dfrac{5}{7}=\dfrac{8}{13}.\dfrac{7}{5}=\dfrac{56}{65}< 1\)

a: \(\dfrac{-13}{40}< \dfrac{-12}{40}\)

\(\dfrac{-5}{6}>\dfrac{-91}{104}\)

Sao bn giống BT mình thế ?:)

\(\dfrac{-11}{3^7\cdot7^3}=\dfrac{1}{3^7\cdot7^3}\cdot\left(-11\right)\)

\(\dfrac{-78}{3^7\cdot7^4}=\dfrac{-78}{3^7\cdot7^3\cdot7}=\dfrac{1}{3^7\cdot7^3}\cdot\dfrac{-78}{7}\)

mà \(-11>-\dfrac{78}{7}\)

nên \(\dfrac{-11}{3^7\cdot7^3}>\dfrac{-78}{3^7\cdot7^4}\)

\(a,\dfrac{a}{b}>1\Leftrightarrow a>1\cdot b=b\\ \dfrac{a}{b}< 1\Leftrightarrow a< 1\cdot b=b\\ b,\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab+a}{b^2+b}\\ \dfrac{a+1}{b+1}=\dfrac{b\left(a+1\right)}{b\left(b+1\right)}=\dfrac{ab+b}{b^2+b}\\ \forall a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+1}{b+1}\\ \forall a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+1}{b+1}\\ \forall a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+1}{b+1}\)

\(c,\forall a>b\Leftrightarrow\dfrac{a}{b}-1=\dfrac{a-b}{b}>\dfrac{a-b}{b+n}\left(b< b+n;a-b>0\right)=\dfrac{a+n}{b+n}-1\\ \Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a< b\Leftrightarrow1-\dfrac{a}{b}=\dfrac{b-a}{b}>\dfrac{b-a}{b+n}\left(b< b+n;b-a>0\right)=1-\dfrac{a+n}{b+n}\\ \Leftrightarrow1-\dfrac{a}{b}>1-\dfrac{a+n}{b+n}\Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a=b\Leftrightarrow\dfrac{a+n}{b+n}=\dfrac{a}{b}\left(=1\right)\)