#)Giải :

Ta có : \(x=\frac{-9}{8}=-\frac{9}{8}\)

\(y=\frac{-49}{50}=-\frac{49}{50}\)

Vì x có 9 > 8 ( tử số lớn hơn mẫu số ) và y có 49 < 50 ( tử số bé hơn mẫu số )

\(x>y \left(\frac{-9}{8}>\frac{-49}{50}\right)\)

Bài làm

Ta có: \(x=-\frac{9}{8}>1\)

           \(y=-\frac{49}{50}< 1\)

Mà hai phân số này là số nguyên âm

=>\(x=-\frac{9}{8}>y=-\frac{49}{50}\)

 Hay \(x>y\)

# Học tốt #

Ta có:

\(x=\frac{1989}{1990}=1-\frac{1}{1990}\)

\(y=\frac{2011}{2012}=1-\frac{1}{2012}\)

Do \(\frac{1}{1990}>\frac{1}{2012}\)=> \(-\frac{1}{1990}< -\frac{1}{2012}\) => \(1-\frac{1}{1990}< 1-\frac{1}{2012}\)

=> \(x=\frac{1989}{1990}< y=\frac{2010}{2012}\)

Ta có :

x = \(\frac{1989}{1990}\)= 1 - \(\frac{1}{1990}\)

y = \(\frac{2011}{2012}\)= 1 - \(\frac{1}{2012}\)

Do \(\frac{1}{1990}\)> \(\frac{1}{2012}\)=> \(-\)\(\frac{1}{1990}\)< \(-\)\(\frac{1}{2012}\)=> \(1\)\(-\)\(\frac{1}{1990}\)\(< 1-\)\(\frac{1}{2012}\)

=> \(x\)\(=\)\(\frac{1989}{1990}\)\(< y=\)\(\frac{2010}{2012}\)

\(\text{Ta thấy : }2671\)

\(\Rightarrow\frac{267}{268}-\left(\frac{1347}{1343}\right)\)

\(\text{Hay }\frac{267}{-268}>\frac{-1347}{1343}\)

267 < 268 => -267/268 > - 1

1347 > 1343 => -1347 /1343 < -1

Vậy 267/-268 > -1347/1343

oai oai ( *_* )

Tơ ... đau ... hag

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chịu

\(\frac{2008}{2009}=\frac{1995}{1996}\)

Ta có :

1 - 2008/2009 = 2009/2009 - 2008/2009 = 1/2009

1 - 1995/1996 = 1996/1996 - 1995/1996 = 1/1996

Vì 1/2009 < 1/1996 => 2008/2009 > 1995/1996

\(\frac{2008}{2009}=\frac{2009-1}{2009}=1-\frac{1}{2009}\)

\(\frac{1995}{1996}=\frac{1996-1}{1996}=1-\frac{1}{1996}\)

\(2009>1996\Rightarrow\frac{1}{2009}<\frac{1}{1996}\Rightarrow1-\frac{1}{2009}>1-\frac{1}{1996}\Rightarrow\frac{2008}{2009}>\frac{1995}{1996}\)

ta có : 

\(B=\frac{2010+2011}{2011+2012}=\frac{2010}{2011+2012}+\frac{2011}{2011+2012}\)

ta có : \(\frac{2010}{2011}>\frac{2010}{2011+2012}\)

            \(\frac{2011}{2012}>\frac{2011}{2011+2012}\)

=> \(\frac{2010}{2011}+\frac{2011}{2012}>\frac{2010+2011}{2011+2012}\)

hay A>B