Ta có :

A = \(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{64.69}\)

5A = \(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{64.69}\)

5A = \(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{64}-\frac{1}{69}\)

5A = \(\frac{1}{4}-\frac{1}{69}\)

5A = \(\frac{65}{276}\)

A = \(\frac{65}{276}:5\)

A = \(\frac{13}{276}\)

a) Quy đồng pso và tính như bthg (4824829/6350400)

b) Vì 4814819 < 6350400 => A < 1

Ta có \(\left(9+1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)

\(=\frac{1}{8}\left(9-1\right)\left(9+1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)

\(=\frac{1}{8}\left(9^2-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)

cứ như thế

\(=\frac{1}{8}\left(9^{64}-1\right)< 9^{64}-1\)=>đpcm

Bài 1: 

a: -8/12<0<-3/-4

b: -56/24<0<7/3

c: 4/25<1<15/13

=>-4/25>-15/13

Bài 2: 

a: =-60/45=-4/3

b: =4/15-3/2-8/5=8/30-45/30-48/30=-85/30=-17/6

a: -15/37>-25/37

b: -13/21=-26/42

-9/14=-27/42

mà -26>-42

nên -13/21>-9/14

c: -49/-63=7/9

56/80=7/10

=>-49/-63>56/80

d: 3/14=1-11/14

4/15=1-11/15

mà 11/14>11/15

nên 3/14<4/15