Bài 1: 

1: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)

\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)

mà \(17^{19}+1>17^{18}+1\)

nên 17A>17B

hay A>B

2: \(C=\dfrac{98^{99}+98^{10}+1-98^{10}}{98^{89}+1}=98^{10}+\dfrac{1-98^{10}}{98^{89}+1}\)

\(D=\dfrac{98^{98}+98^{10}+1-98^{10}}{98^{88}+1}=98^{10}+\dfrac{1-98^{10}}{98^{88}+1}\)

mà \(98^{89}+1>98^{88}+1\)

nên C>D

a)=

b)>

c)=

Cô làm rồi em nhé:

https://olm.vn/cau-hoi/giup-em-voiii.8161766187032

a,

Ta có:
\(\dfrac{3}{7}=1-\dfrac{4}{7}\)
\(\dfrac{11}{15}=1-\dfrac{4}{15}\)
So sánh phân số \(\dfrac{4}{7}\) và \(\dfrac{4}{15}\)
Vì \(7< 15\) nên \(\dfrac{1}{7}>\dfrac{1}{15}\)
\(\Rightarrow1-\dfrac{4}{7}< 1-\dfrac{4}{15}\)
Vậy \(\dfrac{3}{7}< \dfrac{11}{15}\)

b)

\(\dfrac{-11}{6}< -1< \dfrac{-8}{9}\) nên \(\dfrac{-11}{6}< \dfrac{-8}{9}\)

c) 

\(\dfrac{305}{25}=\dfrac{305:5}{25:5}=\dfrac{61}{5}\)
Ta có: 
Mẫu số chung 2 phân số: 80
\(\dfrac{297}{16}=\dfrac{297*5}{16*5}=\dfrac{1485}{80}\)
\(\dfrac{61}{5}=\dfrac{61*16}{5*16}=\dfrac{976}{80}\)
Vì \(1485>976\) nên\(\dfrac{1485}{80}>\dfrac{976}{80}\)
Vậy \(\dfrac{297}{16}>\dfrac{305}{25}\)

d,

$\frac{-205}{317}=\frac{-205:-1}{317:-1}=\frac{205}{-317}$
Ta có: 
Mẫu số chung 2 phân số: -35187
\(\dfrac{205}{-317}=\dfrac{205*111}{-317*111}=\dfrac{22755}{-35187}\)
\(\dfrac{-83}{111}=\dfrac{-83*-317}{111*-317}=\dfrac{26311}{-35187}\)
Vì \(22755< 26311\) nên\(\dfrac{22755}{-35187}< \dfrac{26311}{-35187}\)
Vậy \(\dfrac{-205}{317}< \dfrac{-83}{111}\)

Câu d, mình làm sai, cho mình sửa lại:

\(\dfrac{-205}{317}=\dfrac{-22755}{35187}\)

\(\dfrac{-83}{111}=\dfrac{-26311}{35187}\)

Vậy là  \(-22755>-26311\) hay \(\dfrac{-205}{317}>\dfrac{-83}{111}\)

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

(Sửa \(cn-bm\rightarrow cn-dm\))

Ta có :

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

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

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

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

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

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

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

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