Lời giải:

$\frac{179}{197}< 1< \frac{971}{917}$

$\frac{183}{184}> 0> \frac{-184}{183}$

a) 179/197 là số bé

    971/917 là số lớn

=>179/197<971/917

b)183/184 là số bé

   -184/-183=184/183 là số lớn

=> 183/184<184/183

a) Do 179/197 < 1

971/917 > 1

=> 179/197 < 971/917

b) Do 183/184 < 1

-184/-183 = 184/183 > 1

=> 183/184 < -184/-183

a,Vì

179/197 < 1

971/917 > 1

=>179/197 <1<971/917

=>179/197<971/917

b, Vì

183/184 < 1

-184/-183 = 184/183 > 1

=>183/184 <1<184/183

=>183/184<-184/183

Chúc bạn học giỏi nha!!!

1/Vì 179/197<1 ; 971/917>1

=>179/197<971/917

2/Vì 183/184<1 ; 184/183>1

=>183/184<184/183

3/Ta có : -3/31=-3*101/31*101=-303/3131

Vì -303>-789 =>-303/3131>-789/3131 =>-3/31>-789/3131

1/>

2/<

3/>

Duyệt lẹ !Duyệt lẹ !Duyệt lẹ !

1) Ta có: \(\frac{179}{197}<1;\frac{971}{917}>1\)

=> \(\frac{179}{197}<1<\frac{971}{917}\)

=> \(\frac{179}{197}<\frac{971}{917}\)

2) Ta có: \(\frac{183}{184}<1;\frac{184}{183}>1\)

=> \(\frac{183}{184}<1<\frac{184}{183}\)

=> \(\frac{183}{184}<\frac{184}{183}\)

Câu 1:

1) 179/197 và 971/917

Ta có:

\(1-\frac{179}{197}=\frac{18}{197}\)

\(1-\frac{971}{917}=\frac{-54}{917}\)

Mà \(\frac{-54}{917}<\frac{18}{197}\)

\(\Rightarrow\frac{971}{917}<\frac{179}{197}\)

Câu 2:

Ta có:

\(1-\frac{183}{184}=\frac{1}{184}\)

\(1-\frac{184}{183}=\frac{-1}{183}\)

Mà:\(\frac{-1}{183}<\frac{1}{184}\)

\(\Rightarrow\frac{184}{183}<\frac{183}{184}\)

\(\dfrac{-178}{179}>-1>\dfrac{-191}{189}\\ \dfrac{127}{129}=1-\dfrac{2}{129};\dfrac{871}{873}=1-\dfrac{2}{873}\\ \dfrac{2}{129}>\dfrac{2}{873}\left(129< 873\right)\Leftrightarrow1-\dfrac{2}{129}< 1-\dfrac{2}{873}\Leftrightarrow\dfrac{127}{129}< \dfrac{871}{873}\)

Ta có:

\(A=\dfrac{196}{197}+\dfrac{197}{198}\)

\(B=\dfrac{196+197}{197+198}\)

\(=\dfrac{196}{197+198}+\dfrac{197}{197+198}\)

Áp dụng tính chất \(\dfrac{a}{b}>\dfrac{a}{b+m}\) ta có:

\(\left\{{}\begin{matrix}\dfrac{196}{197}>\dfrac{196}{197+198}\\\dfrac{197}{198}>\dfrac{197}{197+198}\end{matrix}\right.\)

\(\Rightarrow\dfrac{196}{197}+\dfrac{197}{198}>\dfrac{196}{197+198}+\dfrac{197}{197+198}=\dfrac{196+197}{197+198}\)

Vậy \(A>B\)

2/

a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)

\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)

-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)

b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)

-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)

c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)

\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)

-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)

a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)

c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)

a) 

b) 

c) \(\dfrac{25}{100}=\dfrac{10}{40}\)

a) \(\dfrac{23}{24}< 1\)

\(\dfrac{24}{23}>1\)

\(\Rightarrow\dfrac{23}{24}< \dfrac{24}{23}\)

b) \(\dfrac{4}{21}< \dfrac{4}{20}=\dfrac{1}{5}=\dfrac{6}{30}< \dfrac{6}{29}\)

c) \(\dfrac{6}{7}=1-\dfrac{1}{7}< \dfrac{8}{9}=1-\dfrac{1}{9}\)

d) \(\dfrac{1212}{1313}=\dfrac{12\times101}{13\times101}=\dfrac{12}{13}\)