Ta có: \(17A=17.\left(\frac{17^{2001}+1}{17^{2002}+1}\right)=\frac{17^{2002}+17}{17^{2002}+1}=\frac{17^{2002}+1+16}{17^{2002}+1}=1+\frac{16}{17^{2002}+1}\)

\(17B=17.\left(\frac{17^{2000}+1}{17^{2001}+1}\right)=\frac{17^{2001}+17}{17^{2001}+1}=\frac{17^{2001}+1+16}{17^{2001}+1}=1+\frac{16}{17^{2001}+1}\)

Vì 1 = 1 và 16 = 16 nên so sánh mẫu:

17²⁰⁰² + 1 > 17²⁰⁰¹ + 1

=> \(1+\frac{16}{17^{2002}+1}<1+\frac{16}{17^{2001}+1}\)

=> 17A < 17B

=> A < B.

Ta có:\(17^{2001}>17^{2000},1=1\) Còn \(\frac{1}{17^{2002}},\frac{1}{17^{2001}}\) thì ko quan trọng chúng đều nhỏ hơn 1

Nên A>B

a,Ta có: \(\frac{2000}{2002}< 1< \frac{2002}{2001}\)suy ra \(\frac{-2000}{2002}>\frac{-2002}{2001}\)

b,Ta có: \(\frac{5}{17}>\frac{5}{20}=\frac{1}{4}=\frac{4}{16}\)

Đáp số:\(\frac{-2000}{2002}>\frac{-2002}{2001}\)và \(\frac{5}{17}>\frac{4}{16}\)

Nữ hoàng tháng 5

a.\(\frac{13}{17}\)=1-\(\frac{4}{17}\);    \(\frac{46}{50}\)=1-\(\frac{4}{50}\)

Vì \(\frac{4}{17}\)>\(\frac{4}{50}\)=> 1-\(\frac{4}{17}\)<1-\(\frac{4}{50}\)

Vậy\(\frac{13}{17}\)<\(\frac{46}{50}\)

c.\(\frac{41}{91}\)=1-\(\frac{50}{91}\)=1-\(\frac{500}{910}\);    \(\frac{411}{911}\)=1-\(\frac{500}{911}\)

Vì \(\frac{500}{910}\)>\(\frac{500}{911}\)=>1-\(\frac{500}{910}\)<1-\(\frac{500}{911}\)=>\(\frac{41}{91}\)<\(\frac{411}{911}\)

d. \(\frac{2001}{2002}< \frac{2002}{2002}=1;\frac{2005}{2003}>\frac{2003}{2003}=1\text{ hay }\frac{2001}{2002}< 1< \frac{2005}{2003}\)

Vậy \(\frac{2001}{2002}< \frac{2005}{2003}\).

e. \(-\frac{2005}{2010}< 0;\frac{2001}{2002}>0\text{ hay }-\frac{2005}{2010}< 0< \frac{2001}{2002}\)

Vậy \(-\frac{2005}{2010}< \frac{2001}{2002}\).

b. \(\frac{33}{131}>\frac{33}{132}=\frac{1}{4};\frac{53}{217}< \frac{53}{212}=\frac{1}{4}\text{ hay }\frac{53}{217}< \frac{1}{4}< \frac{33}{131}\)

Vậy \(\frac{53}{217}< \frac{33}{131}\).

a)\(9^{12}=\left(3^2\right)^{12}=3^{24}\)

 \(27^7=\left(3^3\right)^7=3^{21}\)

\(\Rightarrow9^{12}>27^7\)

a) bạn Mạnh làm rồi và đúng

b) Ta có : \(333^{444}=\left(333^4\right)^{111}=\left[\left(3.111\right)^4\right]^{111}=\left[\left(3^4.111^4\right)\right]^{111}=\left(84.111^4\right)^{111}\)

                \(444^{333}=\left(444^3\right)^{111}=\left[\left(4.111\right)^3\right]^{111}=\left[\left(4^3.111^3\right)\right]^{111}=\left(64.111^3\right)^{111}\)

Ta thấy (84.111⁴)¹¹¹ > ( 64.111³)¹¹¹ => 333⁴⁴⁴ > 444³³³

Vậy...

c) Vì \(17^{2002}+1>17^{2001}+1\)

\(\Rightarrow\frac{17^{2001}+1}{17^{2002}+1}< \frac{17^{2001}+1}{17^{2001}+1}\)

Ta có: B = \(\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}=\frac{2000}{4003}+\frac{2001}{4003}\)

Ta thấy : \(\frac{2000}{2001}>\frac{2000}{4003}\)(1)

             \(\frac{2001}{2002}>\frac{2001}{4003}\) (2)

Từ (1) và (2) cộng vế với vế, ta được :

  \(\frac{2000}{2001}+\frac{2001}{2002}>\frac{2000}{4003}+\frac{2001}{4003}\)

hay \(A=\frac{2000}{2001}+\frac{2001}{2002}>B=\frac{2000+2001}{2001+2002}\)

Ta có:

\(A=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{3999.4000}}\)

\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{3999}-\frac{1}{4000}}\)

\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{3}+...+\frac{1}{3999}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)

\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)

\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2000}\right)}\)

\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}=1\)

Ta lại có: 

\(B=\frac{\left(17+1\right)\left(\frac{17}{2}+1\right)...\left(\frac{17}{19}+1\right)}{\left(1+\frac{19}{17}\right)\left(1+\frac{19}{16}\right)...\left(1+19\right)}\)

\(=\frac{\frac{18}{1}.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{\frac{36}{17}.\frac{35}{16}.\frac{34}{15}...\frac{20}{1}}\)

\(=\frac{1.2.3...36}{1.2.3...36}=1\)

Từ đây ta suy ra được

\(A-B=1-1=0\)

BAN  CO THE TINH RO BIEU THUC B KO?

xin lỗi, mk mới học lớp 5 nên ko giúp đc gì, mong cậu thông cảm, chúc cậu học giỏi nha

ta có:\(A=\frac{2000}{2001}+\frac{2001}{2002}<\frac{2000}{2002}+\frac{2001}{2002}=\frac{2000+2001}{2002}<\frac{2000+2001}{2001+2002}=B\)

\(\Rightarrow A

ta có:\(B=\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)

vì \(\frac{2000}{2001}>\frac{2000}{2001+2002}và\frac{2001}{2002}>\frac{2001}{2001+2002}\)

\(\Rightarrow\frac{2000}{2001}+\frac{2001}{2002}>\frac{2000+2001}{2001+2002}\)

=>A>B

A = \(\frac{2000+2001}{2001+2002}\)= \(\frac{4001}{4003}\)

B = \(\frac{2000+2001}{2001+2003}=\frac{4001}{4003}\)

vậy A = B

A=B chứ còn gì

$A=\frac{2000+2001}{2001+2002}$A=2000+20012001+2002 

$B=\frac{2000+2001}{2001+2002}$B=2000+20012001+2002 

=>A=B

a) \(\frac{3}{-4}=\frac{-3}{4};\frac{-1}{-4}=\frac{1}{4}\)

Vì - 3 < 1 nên \(\frac{-3}{4}< \frac{1}{4}\)

hay \(\frac{3}{-4}< \frac{-1}{-4}\)

Quy đồng mẫu ta được:

15/17=15.27/17.27=405/459

25/27=25.17/27.27=425/459

⇒405/459<425/459⇒15/17<25/27

c) quy đồng

-9/6=-36/24

6/-4=36/-24

tương đương -36/24=-36/24

suy ra -9/6=6/-4