dài thế bị rảnh à

type nhanh mà bạn

dài z mà làm ak

A=\(\frac{3^{10}+1}{3^9+1}\)>1  

=> A=\(\frac{3^{10}+1}{3^9+1}\)> \(\frac{3^{10}+1+2}{3^9+1+2}\)

=>A=\(\frac{3^{10}+1}{3^9+1}\)>\(\frac{3^{10}+3}{3^9+3}\)

=>A=\(\frac{3^{10}+1}{3^9+1}\)>\(\frac{3\left(3^9+1\right)}{3.\left(3^8+1\right)}\)

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

vậy A>B

vì 1 là bạn quy đồng lên 

 hai là giải thich như sau

 \(\frac{25}{29}\)= 1- \(\frac{4}{29}\)

\(\frac{31}{35}\)= 1- \(\frac{4}{35}\)   mà \(\frac{4}{29}\)> \(\frac{4}{35}\)

nên 1-\(\frac{4}{29}\)<1-\(\frac{4}{35}\)

vậy \(\frac{25}{29}\)<\(\frac{31}{35}\)

a. \(\frac{-2525}{2929}\)= -\(\frac{25.101}{29.101}\)=\(\frac{-25}{29}\)

\(\frac{-217}{245}\)=\(\frac{-7.31}{7.35}\)=\(\frac{-31}{35}\)

vì \(\frac{25}{29}\)>\(\frac{31}{35}\) => \(\frac{-25}{29}\)<\(\frac{-31}{35}\)

b.ta có 27.75=2025

26.82=2132

=> 26.82> 27.75

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

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

=>

2                                                                

2                            

2 nha bạn

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

\(\Rightarrow A< B\)

ai trả lời nhanh nhất mk cho 2 k!!!

a. Ta có :

\(\frac{1}{2}< \frac{1}{3}\)

\(\frac{2}{3}>\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{3}< \frac{1}{2}< \frac{2}{3}\)