a, Ta có: \(\dfrac{27}{37}< \dfrac{27}{18};\dfrac{27}{18}< \dfrac{28}{18}\Rightarrow\dfrac{27}{37}< \dfrac{28}{18}\)

b, Ta có: \(1-\dfrac{2003}{2005}=\dfrac{2}{2005}\)

\(1-\dfrac{2001}{2003}=\dfrac{2}{2003}\)

Vì \(\dfrac{2}{2005}< \dfrac{2}{2003}\Rightarrow\dfrac{2003}{2005}>\dfrac{2001}{2003}\)

Ta thấy \(\frac{2019}{2001}< 1\)và \(\frac{2017}{2003}>1\)

\(=>\frac{2019}{2001}< \frac{2017}{2003}\)

~ Chúc hok tốt ~

\(\frac{2019}{2001}=2019:2001\)

\(\frac{2017}{2003}=2017:2003\)

ta có 2019:2001=1,008995502 và 2017:2003=1,006989516

ta có 1,008995502>1,006989516

\(\Rightarrow\)\(\frac{2019}{2001}>\frac{2017}{2003}\)

CHÚC BẠN HỌC TỐT

ta có 2019/2001=2019:2001 và 2017/2003=2017:2003

vì 2019:2001 >2017:2003

nên 2019/2001 >2017/2003

A = 2001 x 2005                       

= 2001 x( 2003 + 2)                   

= 2001 x 2003 + 2001 x 2          

B = 2003 x 2003

    = (2001+2)x 2003

    = 2001 x 2003 + 2003 x 2

  Vì  2001 x 2 <    2003  2 nên A < B

2001/2001<2005/2003 vì 2001/2001=1 mà 2005/2003 >1 Vậy 2001/2001<2005/2003

Có \(\frac{2001}{2001}=1=\frac{2003}{2003}\)

Mà \(\frac{2003}{2003}< \frac{2005}{2003}\)

=>\(\frac{2001}{2001}< \frac{2005}{2003}\)

A= 2001*2005

A=2001*(2003+2)

A=2001*2003+2001*2

A=2001*2003+4002

B=2003*2003

B=(2001+2)*2003

B=2001*2003+2*2003

B=2001*2003+4006

vi 4006 >4002

=> 2001*2003+4006>2001*2003+4002

=>B>A 

vậy B>A

ko biet luon a bn

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\(\dfrac{2001+2002}{2002+2003}< \dfrac{2001}{2002}+\dfrac{2002}{2003}\)