so = nhau ban oi

A=\(\frac{1999x10000001}{2000x10000001}\) A=\(\frac{1999}{2000}\)

 Vì \(\frac{1999}{2000}\)=\(\frac{1999}{2000}\)

=>A=B

\(B=\frac{1999+2000}{2000+2001}\)

\(B=\frac{1999}{2000+2001}+\frac{2000}{2000+2001}\)

Vì \(\frac{1999}{2000+2001}< \frac{1999}{2000}\) ; \(\frac{2000}{2000+2001}< \frac{2000}{2001}\)

\(\Rightarrow\)\(B=\frac{1999}{2000+2001}+\frac{2000}{2000+2001}\)<  \(A=\frac{1999}{2000}+\frac{2000}{2001}\)

\(\Rightarrow\)B < A

Vậy B < A

ta thấy 1999¹⁹⁹⁹ + 1 / 1999²⁰⁰⁰ + 1 < 1 và 1998 > 0

nên ta có: A < 1999¹⁹⁹⁹ + 1 + 1998 / 1999²⁰⁰⁰ + 1 + 1998

                    < 1999¹⁹⁹⁹ + 1999 / 1999²⁰⁰⁰ + 1999

                    < 1999(1999¹⁹⁹⁸ + 1) / 1999(1999¹⁹⁹⁹ + 1)

                    < 1999¹⁹⁹⁸  + 1 / 1999¹⁹⁹⁹ + 1 

                    < B

Vậy A < B

để tui xem lại đã hink như tui làm bài này zùi

a: 1974/1975<1<2000/1999

b: 2005/2005=1>1993/1999

a)1999/2001<1

12/11>1

=>1999/2001<12/11

b)

1998/1999=1-1/1999

1999/2000=1-1/2000

Vì 1/1999>1/2000

=>1998/1999<1999/2000

A) 1 - 1974/1975 = 1/1975

1 - 1998/1999 = 1/ 1999

Mặt khác 1/1975 > 1/1999

=> 1974/1975 < 1998/1999

Ta có 1974/1975 =1-1974/1975=1975/1975-1

974/1975=1/1979.

1998/1999 =1-1998/1999=1999/1999-1998/1999 =1/1999. 

Vì 1/1979 l/h 1/1999 nên 1974 /1975 l/h 1998/1999.

B = \(\frac{2001}{2002}+\frac{2002}{2003}\)

có: \(\frac{2000}{2001}>\frac{2000}{2001}+2002\)

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

Vậy A>B

Ta có A = 1999 x 2001

             = ( 2000 - 1) x ( 2000 + 1)

              = 2000² -1

Mà B = 2000 x 2000 = 2000²

=> A < B

A = 1999 x ( 2000 + 1) = 1999 x 2000 + 1999

B = (1999 + 1) x 2000 = 1999 x 2000 + 2000

A < B