a) ta có:  \(1-\frac{2012}{2013}=\frac{1}{2013}\)

                 \(1-\frac{2013}{2014}=\frac{1}{2014}\)

mà \(\frac{1}{2013}>\frac{1}{2014}\) nên   \(\frac{2013}{2014}>\frac{2012}{2013}\)

sao giống lớp 4 thế ta

\(\frac{2}{7}\)\(< \)\(\frac{4}{9}\)

2/7<4/9

\(\frac{2}{7}< \frac{4}{9}\)

Tk mk mk tk lại ! ^^ 

4/9<1/2

4/9<1/2

Bài 1 :

a) \(\dfrac{12}{15}< \dfrac{12}{14}< \dfrac{13}{14}\Rightarrow\dfrac{12}{15}< \dfrac{13}{14}\)

b) \(\dfrac{11}{12}< \dfrac{11+1984}{12+1984}=\dfrac{1995}{1996}\)

\(\Rightarrow\dfrac{11}{12}< \dfrac{1995}{1996}\)

c) \(\dfrac{499}{498}>\dfrac{499+1}{498+1}=\dfrac{500}{499}\)

\(\Rightarrow\dfrac{499}{498}>\dfrac{500}{499}\)

d) \(\dfrac{51}{80}< \dfrac{51}{79}< \dfrac{53}{79}\)

\(\Rightarrow\dfrac{51}{80}< \dfrac{53}{79}\)

Bài 2:

a) \(\dfrac{1}{9\times10}+\dfrac{1}{10\times11}+\dfrac{1}{11\times12}+...+\dfrac{1}{29\times30}\)

\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+...+\dfrac{1}{29}-\dfrac{1}{30}\)

\(=\dfrac{1}{9}-\dfrac{1}{30}\)

\(=\dfrac{10}{90}-\dfrac{3}{90}\)

\(=\dfrac{7}{90}\)

b) \(\dfrac{2}{7\times9}+\dfrac{2}{9\times11}+\dfrac{2}{11\times13}+....+\dfrac{2}{37\times39}\)

\(=\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{37}-\dfrac{1}{39}\)

\(=\dfrac{1}{7}-\dfrac{1}{39}\)

\(=\dfrac{32}{273}\)

c) \(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+...+\dfrac{1}{31\times34}\)

\(=\dfrac{1}{3}\cdot\left[3\cdot\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+...+\dfrac{1}{31\times34}\right)\right]\)

\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+...+\dfrac{3}{31\times34}\right)\)

\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)\)

\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{34}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{33}{34}\)

\(=\dfrac{11}{34}\)

1.b.1995/1996 và 11/12 = 1 - 1995/1996 và 1 - 11/12 = 1/1996 và 1/12

=1/1996 < 1/12 vậy 1995/1996 < 11/12

c.500/499 và 499/498 = 1 + 500/499 và 1 + 499/498 = 1/499 và 1/498

=1/499 < 1/498 vậy 500/499 < 499/498

2.a.1/9 x 10 + 1/10 x 11 + 1/11 x 12 + ... + 1/29 x 30

= 1/9 - 1/10 + 1/10 - 1/11 + 1/11 - 1/12 + ... + 1/29 - 1/30

=1/9 - 1/30 = 7/90

b.2/7 x  9 + 2/9 x 11 + 2/11 x 13 + ... + 2/37 x 39

=2 x (1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 + ... + 1/37 - 1/39)

=2 x (1/7 - 1/39)

=64/275

c.1/1 x 4 + 1/4 x 7 + 1/7 x 10 + ... + 1/31 x 34

=1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/31 - 1/34

=1/1 - 1/34

=1/34

- 3/4 < 4/3 (vì 3/4 < 1 < 4/3 nên 3/4 < 4/3).

- 10/7 > 7/9 (vì 10/7 > 1 > 7/9 nên 10/7 > 7/9).

- 12/11 > 11/10 (vì 12/11 > 1 > 11/10 nên 12/11 > 11/10).

3/4 < 4/3 vi 3/4 be hon 1con 4/3 lon hon 1

10/7 > 7/9 vi 10/7 lon hon 1 con 7/9 be hon 1

12/11 > 11/10 vi 12/11 lon hon 1 con 11/10 be hon 1

Lời giải:
$\frac{1234}{1334}=1-\frac{100}{1334}> 1-\frac{100}{1330}=1-\frac{10}{133}=\frac{123}{133}$

Ta rút gọn 2 phần số thì được:

\(\frac{3636363636}{2424242424}\) =  \(\frac{36}{24}\) = \(\frac{3}{2}\)

\(\frac{454545454545}{606060606060}\) = \(\frac{45}{60}\) = \(\frac{3}{4}\)

Vậy ta so sánh:

\(\frac{3}{2}\) >   \(\frac{3}{4}\) 

Nên ..

\(\frac{3636363636}{2424242424}\) >  \(\frac{454545454545}{606060606060}\)

~ Hok T ~