Bài 3: Tìm số nguyên n để B = \(\dfrac{2n+9}{n+4}\) là số nguyên.

f)\(\dfrac{12}{47}\) và \(\dfrac{19}{77}\)

Ta có :\(\dfrac{12}{47}< \dfrac{12}{46}=\dfrac{6}{23};\dfrac{19}{77}< \dfrac{19}{76}=\dfrac{1}{4}\)

Vậy \(\dfrac{12}{47}< \dfrac{19}{77}\)

g)\(\dfrac{-58}{89}\) và \(\dfrac{-36}{53}\)

Ta có :\(\dfrac{-58}{89}< \dfrac{-58}{88}=\dfrac{-29}{44};\dfrac{-36}{53}>\dfrac{-36}{52}=\dfrac{-9}{13}\)

Vậy \(\dfrac{-58}{89}< \dfrac{-36}{53}\)

h)\(\dfrac{n+1}{n+2}\) và \(\dfrac{n}{n+3}\)

Ta có : \(\dfrac{n+1}{n+2}=\dfrac{n}{n+\left(2+1\right)}\) và \(\dfrac{n}{n+3}\)

\(\dfrac{n}{n+3}=\dfrac{n}{n+3}\)

Vậy \(\dfrac{n+1}{n+2}=\dfrac{n}{n+3}\)

d)\(2003\cdot2004-\dfrac{1}{2003}\cdot2004\) và \(2004\cdot2005-\dfrac{1}{2004}\cdot2005\)

\(=2004-\left(2003\cdot\dfrac{1}{2003}\right)\) và \(2005-\left(2004\cdot\dfrac{1}{2004}\right)\)

\(=2004-1\) và \(2005-1\)

=2003 và 2004

Vì 2003<2004 nên \(2003\cdot2004-\dfrac{1}{2003}\cdot2004< 2004\cdot2005-\dfrac{1}{2004}\cdot2005\)

e)\(\dfrac{18}{31}\) và \(\dfrac{15}{37}\)

Ta có \(\dfrac{18}{31}< \dfrac{18}{30}=\dfrac{3}{5};\dfrac{15}{37}< \dfrac{15}{36}=\dfrac{5}{12}\)

Vậy \(\dfrac{18}{31}>\dfrac{15}{37}\)

a)\(\dfrac{77}{76}\) và \(\dfrac{84}{83}\)

Ta có : \(\dfrac{77}{76}< \dfrac{77}{75};\dfrac{84}{83}>\dfrac{84}{82}=\dfrac{42}{41}\)

Vậy \(\dfrac{77}{76}< \dfrac{84}{83}\) .

b)\(\dfrac{456}{461}\) và \(\dfrac{123}{128}\)

Ta có : \(\dfrac{456}{461}< \dfrac{456}{460}=\dfrac{114}{115};\dfrac{123}{128}>\dfrac{123}{126}=\dfrac{41}{42}\)

Vậy \(\dfrac{456}{461}>\dfrac{123}{128} \)

c)\(54\cdot107-\dfrac{53}{53}\cdot107+54\) và \(135\cdot269-\dfrac{133}{134}\cdot269+135\)

=\(54\cdot107-107+54\) và \(135\cdot269-\dfrac{133}{134}\cdot209+135\)

=\(54+54\) và \(135-269\cdot\left(\dfrac{133}{134}+1\right)\)
\(=\dfrac{108}{1}\) và \(\left(-134\right)\cdot\dfrac{267}{134}\)

= 108 và (-267)

Vì 108 >-267 nên \(54\cdot107-\dfrac{53}{53}\cdot107+54>135\cdot269-\dfrac{133}{134}\cdot1269+135\) .