a/ Ta có:

\(\frac{3}{124}=\frac{30}{1240}\) ; \(\frac{1}{41}=\frac{30}{1230}\) ; \(\frac{5}{207}=\frac{30}{1242}\) ; \(\frac{2}{83}=\frac{30}{1245}\)

Vì các phân số trên đều cùng tử nên ta so sánh mẫu : 1230<1240<1242<1242

                                                  => \(\frac{30}{1230}>\frac{30}{1240}>\frac{30}{1242}>\frac{30}{1245}\)

                                                  Hay : \(\frac{1}{41}>\frac{3}{124}>\frac{5}{207}>\frac{2}{83}\)

b/ Ta có: 

   \(\frac{16}{9}=\frac{48}{27};\frac{24}{13}=\frac{48}{26}\)

Vì 27>26

=> \(\frac{16}{9}< \frac{24}{13}\)

; ; ;

Vì các phân số trên đều cùng tử nên ta so sánh mẫu : 1230<1240<1242<1242

=>

Hay :

b/ Ta có:

Vì 27>26

=>

Ta có : \(A=\frac{3^{10}+1}{3^9+1}\) => \(A.\frac{1}{3}=\frac{3^{10}+1}{3^{10}+3}=\frac{\left(3^{10}+3\right)-2}{3^{10}+3}=1-\frac{2}{3^{10}+3}\)

           \(B.\frac{1}{3}=\frac{3^9+1}{3^8+1}\Rightarrow B.\frac{1}{3}=\frac{3^9+1}{3^9+3}=\frac{\left(3^9+3\right)-2}{3^9+3}=1-\frac{2}{3^9+3}\)

Vì : \(\frac{2}{3^{10}+3}< \frac{2}{3^9+3}\) nên \(A>B\)

\(a,\frac{27}{82}< \frac{27}{83}=\frac{1}{3};\frac{26}{75}>\frac{25}{75}=\frac{1}{3}\)

nên\(\frac{27}{82}< \frac{26}{75}\)

\(b,\frac{49}{78}< \frac{52}{78}=\frac{2}{3};\frac{64}{95}>\frac{64}{96}=\frac{2}{3}\)

nên\(\frac{49}{78}< \frac{64}{95}\Rightarrow\frac{-49}{78}>\frac{64}{-95}\)

c, Rút gọn:\(\frac{2525}{2929}=\frac{25}{29};\frac{217}{245}=\frac{31}{35}\)

Ta có:\(1-\frac{25}{29}=\frac{4}{29};1-\frac{31}{35}=\frac{4}{35}\Rightarrow1-\frac{25}{29}>1-\frac{31}{35}\)

\(\Rightarrow\frac{25}{29}< \frac{31}{35}\)hay\(\frac{2525}{2929}< \frac{217}{245}\)

\(d,A=\frac{3^{10}+1}{3^9+1}=1+\frac{3}{3^9+1}\);\(B=\frac{3^9+1}{3^8+1}=1+\frac{3}{3^8+1}\)

Dễ dàng nhận thấy \(\frac{3}{3^9+1}< \frac{3}{3^8+1}\Rightarrow A< B\)

Xin lỗi bạn e, mk ko làm được. Chúc bạn học tốt

a)\(\frac{27}{82}< \frac{27}{81}=\frac{1}{3}=\frac{26}{78}< \frac{26}{75}\)

\(\Rightarrow\frac{27}{82}< \frac{26}{75}\)

d)\(\frac{3^{10}+1}{3^9+1}< \frac{3^{10}+1+2}{3^9+1}=\frac{3^{10}+3}{3^9+1}=\frac{3.\left(3^9+1\right)}{3.\left(3^8+1\right)}=\frac{3^9+1}{3^8+1}=B\)

\(\Rightarrow A< B\)

1                                                                                    Bài làm

a) Ta có :27/ 82 = 0,329...

              26 /75= 0,346...

Vì 0,329...<0,346... nên 27/82<26/75

b) Ta có  134/3 = 44,66...

               55/21 = 2,61...

               74/19= 3,89...

               116/37= 3.15.....

Vì 44,66.. > 3,89...>3,15...>2,61 nên 134/3>74/19>116/37>55>21

c) Ta có 16/9 : 24/13 = 0,96.....

mà 0,96<1 nên 16/9<24,13

d) Ta có -49/78= -0,628...

             64:-95 = -0,673...

Vì -0,628...> - 0,673... nên -49/78>64:-95

a) 27/82 < 26/75 ( 2025/6250 < 2132\6250)

b) -49/78 > 64/ -95 ( - 3136/7410 > -4992/7410)

c) ta có: \(A=\frac{54.107-53}{53.107}=\frac{53.107+(107-53)}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)

\(B=\frac{135.269-133}{134.269+135}=\frac{134.269+\left(269-133\right)}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)

\(\Rightarrow A< B\)

d) ta có: \(A=\frac{3^{10}+1}{3^9+1}=\frac{3.\left(3^9+1\right)-2}{3^9+1}=\frac{3.\left(3^9+1\right)}{3^9+1}-\frac{2}{3^9+1}=3-\frac{2}{3^9+1}\)

\(B=\frac{3^9+1}{3^8+1}=\frac{3.\left(3^8+1\right)-2}{3^8+1}=\frac{3.\left(3^8+1\right)}{3^8+1}-\frac{2}{3^8+1}=3-\frac{2}{3^8+1}\)

mà \(\frac{2}{3^9+1}< \frac{2}{3^8+1}\Rightarrow3-\frac{2}{3^9+1}< 3-\frac{2}{3^8+1}\)

=> A < B

Cảm ơn Công chúa Ori nhìu nha

Ta có -49.(-95) = 4655 < 4992 = 64.78

=> \(\frac{-49}{78}\)< \(\frac{64}{-95}\)

Ta có 26/75>25/75=1/3 mà 27/82<1/3

=>27/82<26/75

a) Ta có:

+) \(\dfrac{1}{2}=\dfrac{3}{6}\)

+) \(\dfrac{1}{3}=\dfrac{2}{6}\)

+) \(\dfrac{2}{3}=\dfrac{4}{6}\)

=> \(\dfrac{2}{6}< \dfrac{3}{6}< \dfrac{4}{6}\)

hay \(\dfrac{1}{3}< \dfrac{1}{2}< \dfrac{2}{3}\)

b) Ta có:

+) \(\dfrac{4}{9}=\dfrac{56}{126}\)

+) \(-\dfrac{1}{2}=-\dfrac{63}{126}\)

+) \(\dfrac{3}{7}=\dfrac{54}{126}\)

=> \(-\dfrac{63}{126}< \dfrac{54}{126}< \dfrac{56}{126}\)

hay \(-\dfrac{1}{2}< \dfrac{3}{7}< \dfrac{4}{9}\)

c) Ta có:

+) \(\dfrac{27}{82}=\dfrac{2025}{6150}\)

+) \(\dfrac{26}{75}=\dfrac{2132}{6150}\)

=> \(\dfrac{2025}{6150}< \dfrac{2132}{6150}\)

hay \(\dfrac{27}{82}< \dfrac{26}{75}\)

d) Ta có:

+) \(-\dfrac{49}{78}=-\dfrac{4655}{7410}\)

+) \(-\dfrac{64}{95}=-\dfrac{4992}{7410}\)

=> \(-\dfrac{4665}{7410}>-\dfrac{4992}{7410}\)

hay \(-\dfrac{49}{78}>-\dfrac{64}{95}\)